Question

Hydrogen $\left({ }_{1} H ^{1}\right)$, Deuterium $\left({ }_{1} H ^{2}\right)$, singly ionised Helium $\left({ }_{2} He ^{4}\right)^{+}$ and doubly ionised lithium $\left({ }_{3} Li ^{6}\right)^{++}$ all have one electron around the nucleus. Consider an electron transition from $n=2$ to $n=1$. If the wave lengths of emitted radiation are $\lambda_{1}, \lambda_{2^{\prime}} \lambda_{3}$ and $\lambda_{4}$ respectively then approximately which one of the following is correct?

• A. $4\lambda_1=2\lambda_2,=2\lambda_3=\lambda_4$
• B. $\lambda_1=2\lambda_2,=2\lambda_3=\lambda_4$
• C. $\lambda_1=\lambda_2,=4\lambda_3=9\lambda_4$
• D. $\lambda_1=2\lambda_2,=3\lambda_3=4\lambda_4$

Answer 1

Answer: (C)

$\frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n^{2}}\right] $
$\Rightarrow \lambda \propto \frac{1}{Z^{2}} \text { for given } n_{1}$ & $n_{2} $
$\Rightarrow \lambda_{1}=\lambda_{2}=4 \lambda_{3}=9 \lambda_{4}$