Modern Physics Ques 18
- In a hydrogen like atom, when an electron jumps from the $M$-shell to the $L$-shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be (Main 2019, 11 Jan II)
(a) $\frac{27}{20} \lambda$
(b) $\frac{25}{16} \lambda$
(c) $\frac{20}{27} \lambda$
(d) $\frac{16}{25} \lambda$
Show Answer
Solution:
Formula:
Wavelength Corresponding To Spectral Lines:
- For hydrogen or hydrogen like atoms, we know that
$$ \frac{1}{\lambda}=R Z^{2} \frac{1}{n _1^{2}}-\frac{1}{n _2^{2}} $$
where, $R$ is Rydberg constant and $Z$ is atomic number. When electron jumps from $M$ - shell to the $L$ - shell, then
$\therefore$ Eq (i) becomes
$$ \begin{aligned} & n _1=2 \\ & n _2=3 \end{aligned} $$
(for $L$ - shell)
(for $M$ - shell)
$$ \left.\frac{1}{\lambda}=R Z^{2} \frac{1}{2^{2}}-\frac{1}{3^{2}}=\frac{5}{36} R Z^{2}\right] $$
Now, electron jumps from $N$-shell to the $L$ - shell, for this
$$ \begin{array}{lr} n _1=2 & \text { (for } L-\text { shell) } \\ n _2=4 & \text { (for } N-\text { shell) } \end{array} $$
$\therefore$ Eq. (i) becomes
$$ \frac{1}{\lambda^{\prime}}=R Z^{2} \frac{1}{2^{2}}-\frac{1}{4^{2}}=\frac{3}{16} R Z^{2} $$
Now, we divide Eq (ii) by Eq (iii)
$$ \begin{aligned} \frac{\lambda^{\prime}}{\lambda} & =\frac{5}{36} R Z^{2} \div \frac{3}{16} R Z^{2}=\frac{20}{27} \\ \text { or } \quad \lambda^{\prime} & =\frac{20}{27} \lambda \end{aligned} $$
BETA
