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When a hydrogen atom in a particularly excited state n returns to the ground state, 6 different photons is emitted. Which of the following is/are incorrect?
A. Out of 6 different photons, only 2 photons have a speed equal to that visible light
B. If the highest energy photon emitted from the above sample is incident on the metal plate having a work function 8eV. KE of liberated photo-electron may be equal to or less than 4.75 eV
C. The total number of radial nodes in all the orbitals of ${n^{th}}$the shell is $14$
D. The total number of angular nodes in all the orbitals in ${(n - 1)^{th}}$ the shell is 13

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Last updated date: 13th Jun 2024
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Answer
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Hint: An atom is said to be excited if it acquires enough energy to leave its stable configuration and jumps to the higher orbital. An electronically excited electron retains its state for only a few nanoseconds.

Complete step by step answer:
When a hydrogen atom in a particularly excited state n returns to the ground state, 6 different photons are emitted. Since the photon is defined as the smallest packet of light energy and the velocity of light is maximum which is equal to $3 \times {10^8}m/\sec $thus it is not possible that photons have different velocity hence the option A is incorrect.
If we talk about the energy of emitted photon it can be calculated by the formula given below:
$\Delta E = 13.6(\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}})eV$ (1)
Where ${n_1}$= lower energy state
${n_2}$= higher energy state
Here given that the number of the photon emitted = 6 and the excited state will be calculated as:

$\dfrac{{n(n - 1)}}{2} = 6$
$ \Rightarrow n(n - 1) = 6 \times 2$
$ \Rightarrow n(n - 1) = 12$
$ \Rightarrow {n^2} - n = 12$

Or

$ \Rightarrow {n^2} - n - 12 = 0$
by factorization we get:
$ \Rightarrow {n^2} + 3n - 4n - 12 = 0$
$ \Rightarrow n(n + 3) - 4(n + 3) = 0$
$ \Rightarrow (n - 4)(n + 3) = 0$
$ \Rightarrow n = 4{\text{ Or }}n = - 3$

Thus the photon having the highest energy will go from ${4^{th}}$ to ${1^{st}}$energy level or we can say here ${n_2} = 4$ and ${n_1} = 1$. On putting the value of energy levels in equation 1 we get the energy of photon as:

$\Delta E = 13.6(\dfrac{1}{{{1^2}}} - \dfrac{1}{{{4^2}}})$
$ \Rightarrow \Delta E = 13.6(\dfrac{1}{1} - \dfrac{1}{{16}})$
$ \Rightarrow \Delta E = 13.6(\dfrac{{16 - 1}}{{16}})$
$ \Rightarrow \Delta E = 13.6 \times \dfrac{{15}}{{16}}$
$ \Rightarrow \Delta E = 12.75eV$

As we know the highest energy emitted is equal to the sum of the work-function of the metal plate and the maximum kinetic energy of the photo-electron liberated from the metal surface. The work function given is 8 eV and the maximum kinetic energy given is 4.75 eV thus the maximum energy liberated is given = $8 + 4.75 = 12.75$. Thus we can say option B is correct.
If we talk about option C, the total number of radial nodes in ${n^{th}}$the shell is equal to:
$6(3 + 2 + 1 + 0) = 6 \times 6 = 36$
And the total number of angular nodes in ${(n - 1)^{th}}$ the shell is $3(2 + 1 + 0) = 9$.

Thus the options C and D are incorrect which is the required answer.

Note: The intermediate region where the probability of finding the electron is zero is known as the nodal surface or nodes. It may be noted that s orbitals have only spherical or radial nodes whereas p and d orbitals have both spherical and angular nodes.