Question

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from $n =4$ to $n =2$ of $He ^{+}$spectrum

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

Answer 1

Answer: (C)

$He ^{+}$ion :
$\frac{1}{\lambda( H )}= R (1)^2\left[\frac{1}{ n _1^2}-\frac{1}{ n _2^2}\right] $
$ \frac{1}{\lambda\left( He ^{+}\right)}= R (2)^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$
Given $\lambda( H )=\lambda\left( He ^{+}\right)$
$ R (1)^2\left[\frac{1}{ n _1^2}-\frac{1}{ n _2^2}\right]= R (4)\left[\frac{1}{2^2}-\frac{1}{4^2}\right] $
$ \frac{1}{ n _1^2}-\frac{1}{ n _2^2}=\frac{1}{1^2}-\frac{1}{2^2}$
On comparing $n _1=1 \& n _2=2$