Question

Hydrogen atoms in a particular excited state 'n', when all returned to ground state, 6 different photons are emitted. which of the following is\are incorrect.

• A. out of 6 different photons only two photons will have speed equal to that visible light.
• B. if highest energy photon emitted from from the above sample is incide on the metal 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 two photons will have speed equal to that visible light.
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.

(A, C, D)
Number of photons emitted = 6
so,$\frac{n(n-1)}{2}$ = 6 $\Rightarrow$n = 4
excited state is $3^{rd}$ or n = 4
photon having highest energy will get 4$\to$ 1
so, its energy will be
= $13.6(\frac{1}{1^2}-\frac{1}{4^2})=13.6\times \frac{15}{16}=12.75eV$


when this photon of highest energy is incident on plate having work function 8 eV then $KE=12.75-8=4.75$, KE will be equal to this value or may be less if electron is inner electron. so option (B) is correct.


Option (a) is incorrect because all photon have equal velocity which is $3\times {10}^{8}m/s$ (speed of light).

We know,
For n=4 possible values of $l$ are 0,1,2,3.
where $l$ is the azimuthal quantum no.

Radial nodes= $\sum n-l-1$
Using all possible values of $l$ and n=4, we get,
Radial nodes= 3+2+1+0=6
Option (c) is incorrect because number of nodes in $n^{th}$ shell are 6 radial nodes,


Angular nodes= $\sum l$
Using all possible values of $l$ for $(n-1{)}^{th}=3^{rd}$, we get,
Angular nodes= 2+1+0=3
Option (d) is incorrect because number of angular nodes in $(n-1{)}^{th}$ shell is 3.