Question

It is observed that some of the spectral lines in hydrogen spectrum have wavelengths almost equal to those of the spectral lines in $H{e}^{+}$ ion, Out of the following the transitions in $H{e}^{+}$ that will make this is :

• A. $n=3$ to $n=1$
• B. $n=6$ to $n=4$
• C. $n=5$ to $n=3$
• D. $n=3$ to $n=2$

Answer 1

The correct option is B $n=6$ to $n=4$

Solution

Let those transitions in H be from

$n_1$ to $n_2$ and in $H{e}^{+}$ be from

$n_1^{\prime }$ to $n_2^{\prime }$ so,

In H, $\frac{1}{h}=R\left(1{)}^2\right[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$ ___ (i)

Where $R\to$ Rydberg's constant.

similarly in $H{e}^{+}$,

$\frac{1}{h}=R\left(2{)}^2\right[\frac{1}{(n_1^1{)}^2}-\frac{1}{(n_2^1{)}^2}]$ ___ (ii)

From (i) and (ii),

$R\left(1{)}^2\right[\frac{1}{n_1^2}-\frac{1}{n_2^2}]=4R[\frac{1}{n_1^1}-\frac{1}{n_2^1{}^2}]$

$\Rightarrow \frac{1}{4}[\frac{1}{n_1^2}-\frac{1}{n_2^2}]=\frac{1}{n_1^1{}^2}-\frac{1}{n_2^1{}^2}$

$\Rightarrow \frac{1}{(2n_1{)}^2}-\frac{1}{(2n_2{)}^2}=\frac{1}{(n_1^1{)}^2}-\frac{1}{(n_2^1{)}^2}$

Thus $n_1^1$ and $n_2^1$ must be even no.

Option - B is correct.