Atomic StructureEasy
Question
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?
Options
A.n = 4 to n = 2
B.n = 4 to n = 1
C.n = 2 to n = 1
D.n = 3 to n = 2
Solution
${\bar{\upsilon}}_{H} = {\bar{\upsilon}}_{He^{+}}$
Or $R \times 1^{2}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right) = R \times 2^{2}\left( \frac{1}{2^{2}} - \frac{1}{4^{2}} \right)$
$\therefore n_{1} = 1\text{ and }n_{2} = 2$
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