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

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A

$n_{2} = 2$ to $n_{1} = 1$

57%

B

$n_{2} = 4$ to $n_{1} = 2$

0%

C

$n_{2} = 3$ to $n_{1} = 1$

36%

D

$n_{2} = 3$ to $n_{1} = 2$

7%

Solution:

$\frac{1}{λ} = R_{H} Z^{2} [\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}]$