Question

What transition in the hydrogen spectrum would have the same wavelength as Balmer transition, $n=4$ to $n=2$ in the $H{e}^{+}$ spectrum?

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

Answer 1

Answer: (D)

$n=2$ to $n=1$

$\frac{1}{\lambda }=R_H{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
$\frac{1}{{\lambda }_{H{e}^{+}}}=R_H\times 2^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$
$=R_H\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{1}{{\lambda }_H}$
for $n_2=2$ to $n_1=1$