# The shortest wavelength of H atom is the Lyman series is lambda1.

Question:

The shortest wavelength of $\mathrm{H}$ atom is the Lyman series is $\lambda_{1}$. The longest wavelength in the Balmer series of $\mathrm{He}^{+}$is :-

1. $\frac{5 \lambda_{1}}{9}$

2. $\frac{27 \lambda_{1}}{5}$

3. $\frac{9 \lambda_{1}}{5}$

4. $\frac{36 \lambda_{1}}{5}$

Correct Option: , 3

Solution:

As we know $\Delta \mathrm{E}=\frac{h \mathrm{c}}{\lambda}$

So $\quad \lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}$ for $\lambda$ minimum i.e.

shortest; $\Delta \mathrm{E}=$ maximum

for Lyman series $\mathrm{n}=1 \&$ for $\Delta \mathrm{E}_{\max }$

Transition must be form $\mathrm{n}=\infty$ to $\mathrm{n}=1$

So $\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \mathrm{Z}^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$

$\frac{1}{\lambda}=R_{H} Z^{2}(1-0)$

$\frac{1}{\lambda}=\mathrm{R} \times(1)^{2} \Rightarrow \lambda_{1}=\frac{1}{\mathrm{R}}$

For longest wavelength $\Delta \mathrm{E}=$ minimum for Balmer series $\mathrm{n}=3$ to $\mathrm{n}=2$ will have $\Delta \mathrm{E}$ minimum

for $\mathrm{He}^{+} \mathrm{Z}=2$

So $\frac{1}{\lambda_{2}}=R_{H} \times Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

$\frac{1}{\lambda_{2}}=\mathrm{R}_{\mathrm{H}} \times 4\left(\frac{1}{4}-\frac{1}{9}\right)$

$\frac{1}{\lambda_{2}}=\mathrm{R}_{\mathrm{H}} \times \frac{5}{9}$

$\lambda_{2}=\lambda_{1} \times \frac{9}{5}$