MCQ
Which transition in the hydrogen atomic spectrum will have the same wavelength as the transition, $n = 4$ to $n = 2$ of $He^+$ spectrum?
- A$n = 4$ to $n = 3$
- B$n = 3$ to $n = 2$
- C$n = 4$ to $n = 2$
- ✓$n = 2$ to $n = 1$
✓
Answer
Correct option: D.
$n = 2$ to $n = 1$
d
For $He^+$ ion, $\frac{1}{\lambda } = {Z^2}R\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]$
For $He^+$ ion, $\frac{1}{\lambda } = {Z^2}R\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]$
${(2)^2}R\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{4^2}}}} \right] = \frac{{3R}}{4}$
For hydrogen atoms, $\frac{1}{\lambda } = R\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]$
$\frac{{3R}}{4} = R\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]\,\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}} = \frac{3}{4}$
$n_1 = 1$ and $n_2 = 2.$
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