Question

What electronic transition in the $H{e}^{+}$ spectrum would have the same wavelength as the first Lymen transition of hydrogen:-

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

Answer 1

The correct option is A n = 4 to n = 2

The first Lymen transition of hydrogen
$\frac{1}{{\lambda }_H}=R\left(1{)}^2\right[\frac{1}{1^2}-\frac{1}{2^2}]$ &
For $H{e}^{+}$, $\frac{1}{{\lambda }_{He}^{+}}=R\left(2{)}^2\right[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
But ${\lambda }_H={\lambda }_{He}^{+}$ [as given in question]
$\therefore$ From above equations $n_1=2\&n_2=4$