MCQ
If the shortest wavelength in Lyman series of hydrogen atom is $A$, then the longest wavelength in Paschen series of $He^+$ is
- A$\frac{{5A}}{9}$
- B$\frac{{9A}}{5}$
- C$\frac{{36A}}{5}$
- ✓$\frac{{36A}}{7}$
${n_1} = 1,\,{n_2} = \infty $
$\frac{1}{\lambda } = R{z^2}\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right)$
$ \Rightarrow \frac{1}{A} = {1^2}\,R\left( {\frac{1}{1} - \frac{1}{\infty }} \right) \Rightarrow \frac{1}{A} = R$
Longest wavelength $= 1^{st}$ line
${n_1} = 3,\,{n_2} = 4$
$\frac{1}{\lambda } = R{z^2}\left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) \Rightarrow \frac{1}{\lambda } = \frac{{R7}}{{36}}$
$R = \frac{1}{A}$
$\frac{1}{\lambda } = \frac{{\frac{1}{A} \times 7}}{{36}} \Rightarrow \frac{1}{\lambda } = \frac{7}{{36\,A}} \Rightarrow \lambda = \frac{{36\,A}}{7}$
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