Question

When a transition of electron in $H e^{+}$takes place from $n_{2}$ to $n_{1}$ then wave number in terms of Rydberg constant $R$ will be
( Given $n_{1}+n_{2}=4, n_{2}-n_{1}=2$ )

• A. $\frac{3 R}{4}$
• B. $\frac{8 R}{9}$
• C. $\frac{32 R}{9}$
• D. $\frac{24 R}{9}$

Answer 1

Answer: (C)

$n_{1}+n_{2}=4$
$n_{2}-n_{1}=2$
$2n_{2}=6$
$n_{2}=3$
$n_{1}=1$
$3 \rightarrow 1$
$=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
$=R \times 2^{2}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)$
$=4 R \times \frac{8}{9}$
$=\frac{32 R}{9}$