From Bohr’s Model, Centripetal force of electron = Columbic force of attraction on electron by nucleus.

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Total Nodes=n-1

and

Radial Nodes = n – 1 – ℓ

So, for 3Px n = 3 and ℓ = 1

So, the number of radial nodes = 3 – 1 -1 = 1

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 The yz plane is a two-dimensional plane formed by the y-axis and z-axis. This option suggests that the probability of finding an electron residing in a px orbital is zero when considering positions that lie in this specific plane, because yz plane is the nodal point for px, and at the nodal point the probability of finding an electron is zero._{The yz plane is a two-dimensional plane formed by the y-axis and z-axis. This option suggests that the probability of finding an electron residing in a px orbital is zero when considering positions that lie in this specific plane, because yz plane is the nodal point for px, and at the nodal point the probability of finding an electron is zero.} The yz plane is a two-dimensional plane formed by the y-axis and z-axis. This option suggests that the probability of finding an electron residing in a px orbital is zero when considering positions that lie in this specific plane, because yz plane is the nodal point for px, and at the nodal point the probability of finding an electron is zero.

The maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state is 

2(n−n)(n−n+1) = 25×6 =15 lines.

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Cu(29)=1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰,
l=0⇒S− subshell
Total electrons in S – subshell = 7 

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β- particles will converge while passing through atom as they bear negative charge. if Rutherford had used β- particles instead of α- particles in his Gold-leaf experiment, the observation of β- particles scattering beyond the gold foil would not have been made. This means that the β- particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.

This is the following questions:β- particles will converge while passing through atom as they bear negative charge. if Rutherford had used β- particles instead of α- particles in his Gold-leaf experiment, the observation of β- particles scattering beyond the gold foil would not have been made. This means that the β- particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.

β- particles will converge while passing through atom as they bear negative charge. if Rutherford had used β- particles instead of α- particles in his Gold-leaf experiment, the observation of β- particles scattering beyond the gold foil would not have been made. This means that the β- particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.β- particles will converge while passing through atom as they bear negative charge.

β- particles will converge while passing through atom as they bear negative charge. if Rutherford had used β- particles instead of α- particles in his Gold-leaf experiment, the observation of β- particles scattering beyond the gold foil would not have been made. This means that the β- particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.iment, the observation of β- particles scattering beyond the gold foil would not have been made. This means that the β- particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.

This option suggests that if Rutherford had used β- particles instead of α- particles in his Gold-leaf experiment, the observation of β- particles scattering beyond the gold foil would not have been made. This means that the β- particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.This option suggests that if Rutherford had used β- particles instead of α- particles in his Gold-leaf experiment, the observation of β- particles scattering beyond the gold foil would not have been made. This means that the β- particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.

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n = l is not permissible.
Value of l is always less than the value of n. 

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Fe(Z=26)=[Ar]3d⁶As² 

Fe³⁺ (23 electrons) = [Ar]3d⁵

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The radii of the Bohr orbits are proportional to the square of the principal quantum numbers (n²). The first three Bohr orbits correspond to n = 1, 2, and 3. Therefore, their radii ratio would be 1² : 2² : 3² = 1 : 4 : 9.

Chromium (Cr) with an atomic number of 24 has an electronic configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵. When it loses three electrons to form Cr3+ (3+ charge), its electronic configuration becomes 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³. In this configuration, it has unpaired electrons in the 3d subshell, making it paramagnetic. 

If the angular momentum quantum number of an electron =3, which means (n-1) values are possible for l, so the possible values of l = 0,1,2. The possible values of m will be ± l that are -2,-1,0,1,2, and s will always be +1/2. The value of m will never be more than l, thus this is not a valid set of quantum numbers.  

According to Bohr's model, the speed of an electron in an orbit is inversely proportional to the principal quantum number (n). The formula for the speed (v) of an electron in the nth orbit is given by:  

v ∝ 1/n  

If the speed of an electron in the first orbit (n = 1) is x, then the speed in the third orbit (n = 3) will be: v3 = v1 x (1/n3) = x * (1/3) = x/3

Principal Quantum Number (n): 3 (Third energy level) 

Azimuthal Quantum Number (ℓ): 0 (s orbital) 

Magnetic Quantum Number (mℓ): 0 (One of the orientation states within the s orbital) 

Spin Quantum Number (ms): +1/2 (Spin orientation)

Transition from n = 4 to n = 6 involves the largest energy change, resulting in the absorption of a photon with the lowest frequency and longest wavelength among the given options. This is because the energy of a photon is inversely proportional to its wavelength, and larger energy gaps correspond to lower-frequency transitions.

The de Broglie wavelength (λ) of a particle is given by the formula λ = h / p, where h is the Planck constant and p is the momentum of the particle. For an electron, the momentum can be related to its kinetic energy (K) as p = √(2mK), where m is the mass of the electron. Given the kinetic energy K = 4.55×10-25 J, the mass of an electron (m), and the Planck constant (h), the calculated de Broglie wavelength is indeed 7.28×10-7 m.

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Shortest wavelength means maximum energy. Therefore, the electronic transition involved should be :

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The given electronic configuration is ground state for chromium. Hence, (B) is the correct answer.
 

Fe³+and Mn²+ have same electronic configuration. Hence (B) is the correct answer.  

Principal, azimuthal and magnetic quantum numbers are respectively related to size,shape and orientation
Hence, (B) is the correct answer. 

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Mass of protons is 1.6726 X 10^-27 kg 

Protons are positively charged particles which show deflection towards cathode in electric field 

Positive rays are produced by bombardment of cathode rays on gas molecules in discharge tube.

At low pressure (0.01 torr) there is large spaces between gas molecules and there will be less hinderance for cathode rays to travel from cathode to anode.

Positive rays produce flashes upon ZnS plate

According to Planck's quantum theory, E a v or E = hv

Velocity of photon is independent of wavelength

c= f x λ 

Product of frequency and wavelength remains constant quantity because of inverse relationship between frequency and wavelength

According to Planck's quantum theory, there is no continuous emission or absorption of energy.

Explanation is:

Quantum number which is not derived from Schrodinger wave equation is spin quantum number. It was given by Uhlenbeck and Goudsmith

Nodal plane is the area where the probability of finding of electron is zero

d-subshell has five degenerate orbitals with sausage shape. 

The correct option is b) n = 3, ℓ = 1.

The quantum numbers describe various properties of an electron's orbital in an atom. In the case of the 3p orbital, the correct quantum number values are n = 3 and ℓ = 1.

Let's examine why the other options are incorrect:

a) n = 3, ℓ = 0: This combination represents the 3s orbital, not the 3p orbital. The principal quantum number (n) denotes the energy level or shell, while the azimuthal quantum number (ℓ) indicates the shape of the orbital. The s orbitals have a spherical shape, which corresponds to ℓ = 0.

c) n = 2, ℓ = 1: This combination represents the 2p orbital, not the 3p orbital. The principal quantum number (n) represents the energy level, and in this case, it corresponds to the second shell. However, the 2p orbital belongs to the second shell, not the third shell.

d) n = 2, ℓ = 3: This combination is invalid because the azimuthal quantum number (ℓ) can only have integer values ranging from 0 to (n – 1). Since n = 2, the maximum value of ℓ can be 1, not 3. Therefore, this combination does not correspond to any valid orbital.

To summarize, the correct quantum number values for the 3p orbital are n = 3 and ℓ = 1, as stated in option b.

Note: Number of electrons in a subshell = 2(2ℓ + 1).

The correct option is b) 6.

The p-subshell can accommodate a maximum of 6 electrons. This is because the p-subshell consists of three orbitals (px, py, and pz), and each orbital can hold a maximum of 2 electrons according to the Pauli exclusion principle.

Let's examine why the other options are incorrect:

a) 2 is incorrect because 2 is the maximum number of electrons that can be accommodated in a single p orbital, not the entire p-subshell.

c) 10 is incorrect because 10 is the maximum number of electrons that can be accommodated in a d-subshell, not the p-subshell.

d) 14 is incorrect because 14 is the maximum number of electrons that can be accommodated in an f-subshell, not the p-subshell.

The correct option is a) n=5, ℓ=3, m=+1.

The energy of an electron in an atom depends on its principal quantum number (n), azimuthal quantum number (ℓ), and magnetic quantum number (m). The general trend is that as the principal quantum number increases, the energy of the subshell also increases. Additionally, within a given principal quantum number, the energy increases as the azimuthal quantum number increases.

Let's analyze the other options to understand why they are incorrect:

b) n=5, ℓ=2, m=+2: This subshell has a lower energy compared to option a. The azimuthal quantum number (ℓ) is lower, which indicates a lower energy level. Therefore, option b has a lower energy than option a.

c) n=4, ℓ=3, m=0: This subshell has a lower energy compared to option a. The principal quantum number (n) is lower, which indicates a lower energy level. Therefore, option c has a lower energy than option a.

d) n=4, ℓ=0, m=+1: This subshell has a lower energy compared to option a. The azimuthal quantum number (ℓ) is lower, which indicates a lower energy level. Therefore, option d has a lower energy than option a.

In summary, option a has the highest energy among the given choices because it has the highest principal quantum number (n=5) and highest azimuthal quantum number (ℓ=3).

The correct option is c) 5.

The magnetic quantum number (m) represents the orientation of an orbital within a subshell. For the d-subshell, which has an azimuthal quantum number (ℓ) of 2, the possible values of the magnetic quantum number range from -2 to +2.

Now let's examine why the other options are incorrect:

a) 2: This option is incorrect because the total value of the magnetic quantum number (m) for the d-subshell is 5, not 2.

b) 3: This option is incorrect because the total value of the magnetic quantum number (m) for the d-subshell is 5, not 3.

d) 7: This option is incorrect because the total value of the magnetic quantum number (m) for the d-subshell is 5, not 7.

Therefore, the correct answer is c) 5, as it correctly represents the range of values for the magnetic quantum number in the d-subshell (-2, -1, 0, 1, 2).

Note: Energy of subshell α(n+ ℓ), Energy order s

The correct option is b) Higher. 

An s-orbital has lower energy than a p-orbital of the same principal quantum number.

Lower is incorrect because, as mentioned above, the p-orbital has higher energy than the s-orbital.

Equal is incorrect because the energy of an s-orbital is lower than that of a p-orbital. They are not equal in energy.

Vary orbit to orbit is incorrect because the comparison is between the s-orbital and the p-orbital of the same principal quantum number. While energy levels may vary between different orbitals within the same subshell (such as between px, py, and pz orbitals), the energy of the s-orbital is consistently lower than the energy of the p-orbital.

This difference in energy arises from the different shapes and orientations of the orbitals. The s-orbital is spherically symmetric and has a lower energy, while the p-orbitals have a dumbbell shape and are oriented along the x, y, and z axes, resulting in higher energy.

The correct option is a) n=4, ℓ=0, m=0, s=+½.

The quantum numbers provide specific information about the electron's energy level, orbital shape, orientation, and spin. Let's analyze why the other options are incorrect:

Option b) n=3, ℓ=2, m=0, s=+½: This set of quantum numbers represents the 29th electron being in the 3d orbital. However, copper (Cu) has 29 electrons, and the filling order of electrons follows the Aufbau principle. The 4s orbital is filled before the 3d orbital. Therefore, the 29th electron should occupy the 4s orbital, not the 3d orbital. 

Option c) n=4, ℓ=1, m=0, s=+½: This set of quantum numbers represents the 29th electron being in the 4p orbital. However, as mentioned above, the 29th electron occupies the 4s orbital in copper, not the 4p orbital.

Option d) n=3, ℓ=0, m=0, s=+½: This set of quantum numbers represents the 29th electron being in the 3s orbital. However, copper has 29 electrons, and the 3s orbital can accommodate only 2 electrons. Therefore, the 29th electron cannot be in the 3s orbital.

Hence, option a) n=4, ℓ=0, m=0, s=+½ is correct. It represents the 29th electron being in the 4s orbital of the copper atom.

The correct option is c) dz².

The d-orbital which is bi-lobed with a collar is represented by the dz² orbital.

Option a) dxy: The dxy orbital is characterized by lobes along the x and y axes, without the presence of a collar. It does not exhibit the bi-lobed structure with a collar mentioned in the question.

Option b) dyz: The dyz orbital is characterized by lobes along the y and z axes, without the presence of a collar. It also does not possess the bi-lobed structure with a collar mentioned in the question.

Option d) dz²-y²: The dz²-y² orbital is characterized by two lobes along the z-axis and a nodal plane along the y-axis. It does not exhibit the specific bi-lobed structure with a collar mentioned in the question.

The dz2 orbital, on the other hand, has two lobes along the z-axis and a ring-shaped collar in the xy plane. This collar is responsible for the “bi-lobed with a collar” description mentioned in the question.

The correct option is d) All of these.

Isoelectronic species are atoms or ions that have the same number of electrons. K+ is a potassium ion that has lost one electron, resulting in a configuration of 2, 8, 8. Let's analyze each option:

a) P-3: Phosphorus (P) normally has 15 electrons. By gaining three electrons, it forms the P-3 ion, which has a configuration of 2, 8, 8, 2. This configuration is isoelectronic with K+ (2, 8, 8).

b) Si-4: Silicon (Si) normally has 14 electrons. By gaining four electrons, it forms the Si-4 ion, which has a configuration of 2, 8, 8, 4. This configuration is isoelectronic with K+ (2, 8, 8).

c) S-2: Sulfur (S) normally has 16 electrons. By gaining two electrons, it forms the S-2 ion, which has a configuration of 2, 8, 8, 6. This configuration is isoelectronic with K+ (2, 8, 8).

Therefore, all three options (a, b, and c) are correct because P-3, Si-4, and S-2 are isoelectronic with K+. They all have the same number of electrons as K+ and share the same electron configuration of 2, 8, 8.

So, the correct option is d) All of these.

The correct option is b) 14.

To determine the number of electrons present in the M-shell of an element with an atomic number of 26, we need to determine the electron configuration of the element.

The atomic number 26 corresponds to the element iron (Fe). The electron configuration of iron is:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶

In the ground state, all the lower energy levels (shells) will be filled before the M-shell. Therefore, the M-shell represents the 3rd shell (n=3). The M-shell consists of the 3s, 3p, and 3d subshells.

The 3s subshell can hold a maximum of 2 electrons, the 3p subshell can hold a maximum of 6 electrons, and the 3d subshell can hold a maximum of 10 electrons.

To determine the number of electrons in the M-shell, we need to add up the electrons in the 3s, 3p, and 3d subshells.

3s² + 3p⁶ + 3d⁶ = 2 + 6 + 6 = 14

Therefore, in the M-shell of iron (Fe) in its ground state, there are 14 electrons.

Hence, the correct option is b) 14. The other options (a, c, d) are incorrect because they do not represent the correct number of electrons in the M-shell of iron.

The ground state of an atom is its most stable and lowest energy electron configuration. In this state, electrons fill the available orbitals starting from the lowest energy level (closest to the nucleus) and following the Pauli exclusion principle and Hund's rule. The given electronic configuration “1s², 2s², 2P¹_x, 2P¹_y, 2P¹_z” follows this pattern, with electrons filling the 1s and 2s orbitals completely and then populating the three available 2p orbitals with a total of 5 electrons. This arrangement represents the ground state of a nitrogen atom. 

Electronic configuration of ₂₉Cu = [Ar], 4s¹, 3d¹⁰

Titanium (Ti) has an atomic number of 22. If it loses 4 electrons to become Ti+4, its electronic configuration becomes: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁰ 4s⁰. This configuration is identical to argon's electronic configuration (1s² 2s² 2p⁶ 3s² 3p⁶). Therefore, Ti+4 is the correct answer as it has a similar electronic configuration to argon.

Option C is true for both copper and chromium. Both copper and chromium have electrons in their s-orbitals: Cu has 1 electron in the 4s orbital, and Cr has 1 electron in the 4s orbital.  

Option D violates Hund's rule. According to Hund's rule, electrons preferentially occupy degenerate orbitals singly before pairing up. In this option, the 2p orbitals (2p²_x, 2p⁰_y, 2p⁰_z) should be singly occupied first before pairing, but two electrons are placed in the 2p²_x orbital before any are placed in the 2p⁰_y and 2p⁰_z orbitals. This arrangement violates Hund's rule. 

Half-filled and completely filled arbitas are relatively more stable.
Option B suggests that each of the three electrons is in a different orbital (np¹_x, np¹_Y, np¹_Z). This arrangement follows Hund's rule and is a valid ground state configuration for three electrons in group VA. 

This configuration accurately represents the electronic configuration of phosphorus (P). It has 5 valence electrons in the 3s and 3p orbitals. To form a -3 charged ion, phosphorus would gain 3 electrons, resulting in a stable electron configuration by completing its octet: 1s² 2s² 2p⁶ 3s² 3p⁶.

The electronic configuration of sulfur is [Ne]3s² 3p⁴. The given configuration matches the electron distribution of sulfur (S), with 16 electrons and the distribution 1s² 2s² 2px² 2py² 2pz² 3s² 3px² 3py¹ 3pz¹. Therefore, S is the correct answer. 

Option D suggests that the total number of lobes in all the d orbitals is 18. Each d orbital has a specific number of lobes: dxy, dxz, dyz, and dx²-y² each have 4 lobes, while dz² has 2. Adding these up gives a total of 18 lobes. 

Pauli's exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers, and furthermore, electrons in the same orbital must have opposite spins. This principle is precisely related to the statement given. 

Option A is correct.
If more than one degenerate orbitals are available then electrons go one by one with same spin is called hund's rule.The rule states that electrons go one by one with the same spin in degenerate orbitals and is called Hund's rule of maximum multiplicity. This rule states that electrons will occupy degenerate orbitals singly before they pair up, and that they will have the same spin in each orbital. This is because electrons have a property called spin, which can be either up or down.The Pauli exclusion principle states that no two electrons in an atom can have the same four quantum numbers. This means that if two electrons are in the same orbital, they must have different(opposite) spins. 

Option A is correct.
The e/m value for canal rays is maximum for hydrogen. This is because hydrogen has the lowest mass of all the elements, so the e/m value will be the highest. The other elements you mentioned have higher masses, so their e/m values will be lower. Canal rays are also known as anode rays. They are positively charged ions that are produced when a gas is ionized in a discharge tube. The e/m value of canal rays is a measure of the charge-to-mass ratio of the ions. The higher the e/m value, the lighter the ions are.The e/m value for canal rays is maximum for hydrogen, and it is equal to 1. 

Option C is correct.
The correct order of energy in the given subshells is 5s>3d>4s>3p.
The energy of an electron subshell is determined by two factors: the principal quantum number (n) and the angular momentum quantum number (l). The principal quantum number tells us the energy level of the subshell, and the angular momentum quantum number tells us the shape of the subshell. In general, subshells with higher principal quantum numbers have higher energy than subshells with lower principal quantum numbers. So, 5s has a higher energy than 4s, and 3d has a higher energy than 3p. However, the angular momentum quantum number also plays a role in determining the energy of a subshell. Subshells with the same principal quantum number but different angular momentum quantum numbers have different energies, with subshells with higher angular momentum quantum numbers having higher energy. So, 3d has a higher energy than 3p, even though they have the same principal quantum number.
Therefore, the correct order of energy in the given subshells is 5s>3d>4s>3p. 

Option B is correct.
The maximum number of electrons in a subshell is given by 2(2l + 1). This formula is based on the Pauli exclusion principle, which states that no two electrons in an atom can have the same four quantum numbers. The four quantum numbers are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (ml), and the spin quantum number (ms). The angular momentum quantum number (l) determines the shape of the orbital, and it can have values from 0 to n – 1. The maximum number of electrons in a subshell is therefore 2(2l + 1), where l is the angular momentum quantum number.
No. of orbitals in a subshell = (2ℓ +1)
No. of electrons in a subshell = 2(2ℓ +I)

Option D is correct.
The set of orbitals having n + l = 5 are 5s, 4p, 3d. The principal quantum number (n) tells us the energy level of the orbital, and the angular momentum quantum number (l) tells us the shape of the orbital. The sum of n + l is called the total angular momentum quantum number.
For the 5s orbital, n = 5 and l = 0, so n + l = 5.
For the 4p orbital, n = 4 and l = 1, so n + l = 5.
For the 3d orbital, n = 3 and l = 2, so n + l = 5.
The other sets of orbitals mentioned do not have a total angular momentum quantum number of 5.  

Option A is correct.
The ratio of e/m values of a proton and an alpha particle is 2:1.
The charge of a proton is +1, and its mass is 1 u. The charge of an alpha particle is +2, and its mass is 4 u. The e/m value of a proton is therefore 1/1 = 1, and the e/m value of an alpha particle is 2/4 = 0.5. The ratio of these two values is 2/0.5 = 2:1.
Therefore, the answer is 2:1. 

Option C is correct.
Two electrons present in the same orbital can be distinguished by their spin quantum number. The spin quantum number can have two values, +1/2 and -1/2. This means that there are two possible ways for two electrons to occupy the same orbital, with opposite spins.
The principal quantum number, azimuthal quantum number, and magnetic quantum number all describe the spatial distribution of an electron in an atom. They do not tell us anything about the spin of the electron. Therefore, the answer is spin quantum number.The principal quantum number, azimuthal quantum number, and magnetic quantum number all describe the spatial distribution of an electron in an atom. They do not tell us anything about the spin of the electron. Therefore, the answer is spin quantum number

Option A is correct.
The number of electrons with n = 2, l = 0 in an atom with 35 nucleons and an atomic number of 17 is 2. The principal quantum number (n) tells us the energy level of the electron, and the angular momentum quantum number (l) tells us the shape of the orbital. The value of n = 2 tells us that the electron is in the second energy level, and the value of l = 0 tells us that the orbital is an s orbital. The s orbital in the second energy level can hold up to 2 electrons, so the number of electrons with n = 2, l = 0 in an atom with 35 nucleons and an atomic number of 17 is 2.
Here is the electron configuration of chlorine, which has an atomic number of 17:
1s2 2s2 2p6 3s2 3p5
The electrons with n = 2, l = 0 are the 2s2 electrons. There are 2 of these electrons in chlorine, so the answer is 2. 

Option B is correct.
The e/m ratio of cathode rays is 1.76 x 10^11 C/kg. This means that the charge-to-mass ratio of the particles that make up cathode rays is 1.76 x 10^11 coulombs per kilogram. Cathode rays are beams of electrons that are produced when a high voltage is applied to a gas in a vacuum tube. The electrons are accelerated by the electric field, and they travel in straight lines through the gas. The e/m ratio of cathode rays was first measured by J.J. Thomson in 1897. Thomson's experiment showed that the e/m ratio of cathode rays is independent of the gas that is used in the vacuum tube. This led Thomson to conclude that the particles that make up cathode rays are a fundamental particle of matter, which he called the electron. The e/m ratio of cathode rays is a very important quantity in physics. It is used to determine the mass of the electron, and it is also used to study the properties of electrons. 

Option Ais correct.
The wave number (ν) of a radiation with wavelength λ = 2 x 10^8 nm will be 0.5 x 10^-8 nm^-1. The wave number is defined as the reciprocal of the wavelength. In other words, it is the number of waves per unit distance. The unit for wave number is usually expressed in cm^-1, but it can also be expressed in nm^-1. The wavelength of a wave is the distance between two consecutive peaks or troughs of the wave. In the case of electromagnetic radiation, the wavelength is the distance between two consecutive crests of the wave. The wave number of a radiation can be calculated using the following formula: ν = 1/λ where:
ν is the wave number
λ is the wavelength
In this case, the wavelength is given as 2 x 10^8 nm. So, the wave number can be calculated as follows:
ν = 1/λ = 1/(2 x 10^8 nm) = 0.5 x 10^-8 nm^-1
Therefore, the wave number of a radiation with wavelength λ = 2 x 10^8 nm will be 0.5 x 10^-8 nm^-1.  

According to Aufbau's principle the elcetrons should be filled itno the orbitals according to increasing energy levels. In K electrons first fill into 4s then 3d. Not obeying this rule will change K's block from s to d.

High speed electrons collide with the gas molecules and knock off electrons from them making them positively charged . These ions then travel towards the cathode. Hence, option B is correct.   

 FTB Pg 24      

D is correct beacuse it shows the energy level of electrons , which is often referred as the energy of electron. 

Option D is correct . N is the fourth shell and can accomodate a maximum number of 32e. We can find this by using formula 2n²

Option C is correct , Magnetic quantum number has positive as well as negative value for its quantum number. 

Positive rays are formed when high speed electrons collide with gas molecules and knock off there electrons making them positively charge, but not all gas molecules are ionized. Hence, the ratio of positive rays is less than cathode rays. 

The correct four quantum numbers for an electron in the K-shell are:

1. Principal quantum number (n) = 1
2. Azimuthal quantum number (l) = 0
3. Magnetic quantum number (m) = 0
4. Spin quantum number (s) = +1/2 or -1/2

The principal quantum number (n) determines the energy level and distance from the nucleus, and for the K-shell, it has a value of 1. The azimuthal quantum number (l) determines the shape of the electron's orbital, and for the K-shell, it has a value of 0, indicating that the electron is in an s orbital. The magnetic quantum number (m) specifies the orientation of the orbital in space, and for an s orbital, it can only have a value of 0. The spin quantum number (s) indicates the direction of the electron's spin, and it can have a value of +1/2 or -1/2.

Cathode rays are not electromagnetic in nature, and are made of high speed electrons released from cathode. 

Smaller the size of sub-shell and lower the enrgy level, higher will be the penetration power.

Principal Quantum number shows the orbit is 4 and Azimuthal quantum number 2 shows that electron is in D subshell. So, option C is correct.

Correct option is C

Thus, nitrogen contains three unpaired electrons following Hund's rule of maximum multiplic 

 According to this rule, electron pairing in any orbital (s, p, d or f) does not take place until each orbital of the a subshell contains one electron with paralell spin.
The electronic configuration of nitrogen is 1s2 2s2 2p3.Nitrogen is Trivalent.
FTB Pg42

The shape of an orbital is described by the azimuthal quantum number or orbital angular momentum quantum number (l).

The azimuthal quantum number defines the shape of the orbital and determines the subshell to which the orbital belongs. It specifies the angular momentum of the electron in the orbital and governs the three-dimensional shape of the orbital. 

Spin Quantum Number is not derived from Schrodinger’s wave equation.

correct option is B

The energy of an electron is considered as zero at infinite distance from the nucleus as electron come nearer it gets bounded from the nucleus so the energy is negative. Energy is inversely proportional to the distance so, as the distance decreases energy increases to a greater negative value. 

According to Planck’s quantum theory, The energy of the radiation absorbed or emitted is directly proportional to the frequency of the radiation

The energy of radiation is expressed in terms of frequency as,

E=hν 

Where,

E is the energy of tradition;

h is the Planck’s constant (6.626×10−34 J.s);

ν is the frequency of radiation.  

FTB Pg 35

Fe⁺² has 5 unpaired e^–

The total number of d-electrons in an atom is determined by its atomic number. For an atom with an atomic number of 26, which corresponds to the element iron (Fe), the electron configuration can be determined as follows: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶. 

To find the total number of d-electrons, we look at the electrons in the 3d subshell, which is 6 d-electrons. So, the total number of d-electrons in an atom of atomic number 26 (iron) is 6.
 

Hydrogen ion (H⁺). It contains the same number of electrons as an alpha particle, i.e. 0 electrons. It's worth noting that alpha particles contain the same number of protons as Helium , i.e. 2 protons. There are no electrons in both H⁺ and alpha particle.

The Azimuthal quantum number (l) specifies the shape of an orbital in an atom.The azimuthal quantum number can take on integer values from 0 to (n – 1), where n is the principal quantum number.

However, the azimuthal quantum number (l) cannot be equal to 2 in a 2d orbital because the values of l must follow certain restrictions based on the principal quantum number (n). The allowed values of l for a given n are from 0 to (n – 1).

For example:

• When n = 1, the only allowed value for l is 0.
• When n = 2, the allowed values for l are 0 and 1.
• When n = 3, the allowed values for l are 0, 1 and 2.

In the case of n = 2, the only possible values of l are 0 and 1. These correspond to the s and p orbitals respectively and therefore 2d orbital is not possible.
 

Electronic configuration of Atomic number 24 (Chromium) is :1s²2s²2p⁶3s²3p⁶3d⁵4s¹

M-shell is the third shell, i.e. n = 3

Number of electrons in third shell are 2+6+5 = 13

Hence, the correct option is 13. 

The respective orbital where n = 3, l = 2 will be '3d' because 3d orbitals contain five subshells and the principal quantum number is 3. We know that 3d orbital can accommodate 10 electrons (each subshell can accommodate two electrons). Therefore the number of electrons in n = 3, l = 2 are 10.

Bohr's theory is only applicable to hydrogen-like species containing one electron only. As H, He¹⁺ and Li²⁺ all have one electron each, hence Bohr's theory is applicable on them. But H⁺ has 0 electrons and that is the reason Bohr's theory is not appliable on it.

n=3 represents the third energy level, l=3 indicates an f orbital, m=-3 means the f orbital is oriented in a specific direction, s=+1/2 represents a spin-up electron.

This set of quantum numbers is not possible.

In the third energy level, the maximum value of azimuthal quantum number (l) is 2, so l=3 is invalid. Based on the explanations above, option B (n=3, l=3, m=-3, s=+1/2) is the correct answer as it contains an invalid value for the azimuthal quantum number (l). This option would have been correct if the value of l was smaller than the value of n i.e. less than 3.

Option A is correct. The given element in option A has highest ionization energy as it has half filled p orbital which is comparatively more stable than partially filled p orbital. Hence, greater amount of energy is required to remove an electron from this element. It is extra stable configuration.

Option B is incorrect (refer to the explanation in option B)

Option C is incorrect (refer to the explanation in option C)

Option D is also incorrect (refer to the explanation in option D)