Question

When hydrogen atoms in a particular excited state $n$ returns to the ground state, 6 different photons are emitted. Which of the following is/are incorrect?

• A. Out of $6$ different photons, only $2$ photons have a speed equal to that of visible light
• B. If highest energy photon emitted from the above sample is incident on the matal plate having work function $8eV$. $KE$ of liberated photo-electron may be equal to or less than $4.75eV$
• C. Total number of radial nodes in all the orbitals of $n^{th}$ shell is 14
• D. Total number of angular nodes in all the orbitals in $(n-1{)}^{th}$ shell is 13

Answer 1

Answer: (A), (C), (D)

The correct options are
A Out of $6$ different photons, only $2$ photons have a speed equal to that of visible light
C Total number of angular nodes in all the orbitals in $(n-1{)}^{th}$ shell is 13
D Total number of radial nodes in all the orbitals of $n^{th}$ shell is 14

Number of photons emitted $=6$

So,$\frac{n(n-1)}{2}=6$; $n=4$

Excited state is $3^{rd}$ or $n=4$

Photon having highest energy will go from $4\to 1$

So, its energy will be $=13.6(\frac{1}{1^2}-\frac{1}{4^2})=13.6\times \frac{15}{16}=12.75$ when be equal to this value or may be less if electron is inner electron.

Option (B) is correct because the highest energy photon emitted from the sample (12.75 eV) is equal to the sum of work function of the metal plate (8ev) and maximum kinetic energy of the photoelectron liberated from the metal plate (4.75 eV).

Option (A) is incorrect because all photon have equal velocity which is $3\times {10}^8$ m/s.Radial nodes $=n-l-1$ for a particular orbital.

Angular nodes $=1$ for a particular orbital.

For $n=4$ possible values of $l$ are $0,1,2,3$

Option (C) is incorrect because total number of radial nodes in $n^{th}$ shell are $6(3+2+1+0)$.

Option (D) is incorrect because total number of angular nodes in $(n-1{)}^{th}$ shell are $3(2+1+0)$.