Atoms Ques 13
13. An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be :
[2012]
(a) $\frac{24 h R}{25 m}$
(b) $\frac{25 h R}{24 m}$
(c) $\frac{25 m}{24 h R}$
(d) $\frac{24 m}{25 h R}$
( $m$ is the mass of the electron, $R$, Rydberg constant and $h$ Planck’s constant)
Show Answer
Answer:
Correct Answer: 13.(a)
Solution:
- (a) For emission, the wave number of the radiation is given as
$\frac{1}{\lambda}=R z^{2}(\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}})$
$R=$ Rydberg constant, $Z=$ atomic number
$=R(\frac{1}{1^{2}}-\frac{1}{5^{2}})=R(1-\frac{1}{25}) \Rightarrow \frac{1}{\lambda}=R \frac{24}{25}$
linear momentum
$p=\frac{h}{\lambda}=h \times R \times \frac{24}{25}$ (de-Broglie hypothesis)
$\Rightarrow m v=\frac{24 h R}{25} \Rightarrow v=\frac{24 h R}{25 m}$
BETA
