Question

The shortest wavelength of the H atom in the Lyman series is the longest wavelength in the Balmer series of He⁺ is

• A. $\frac{9{\mathrm{\lambda }}_1}{5}$
• B. $\frac{27{\mathrm{\lambda }}_1}{5}$
• C. $\frac{36{\mathrm{\lambda }}_1}{5}$
• D. $\frac{5{\mathrm{\lambda }}_1}{9}$

Answer 1

The correct option is A

$\frac{9{\mathrm{\lambda }}_1}{5}$

Explanation for the correct answer:-

Option (A) $\frac{9{\mathrm{\lambda }}_1}{5}$

Step 1: Calculating for ${\mathrm{\lambda }}_1$

We know that,

$$\begin{gathered} ∆\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }} \\ \Rightarrow \mathrm{\lambda }=\frac{\mathrm{hc}}{∆\mathrm{E}} \end{gathered}$$

When $\mathrm{\lambda }$ to be minimum that is shortest, then $∆\mathrm{E}=\mathrm{maximum}$

For Lyman series, n=1 and for $∆{\mathrm{E}}_{\max }$

The transition should be from $\mathrm{n}=\infty \mathrm{to}\mathrm{n}=1$

$$\begin{gathered} \therefore \frac{1}{\mathrm{\lambda }}={\mathrm{R}}_{\mathrm{H}}\times {\mathrm{Z}}^2\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right) \\ \Rightarrow \frac{1}{\mathrm{\lambda }}={\mathrm{R}}_{\mathrm{H}}\times {\mathrm{Z}}^2(1-0) \\ \Rightarrow \frac{1}{\mathrm{\lambda }}=\mathrm{R}\times {\left(1\right)}^2 \\ \Rightarrow {\mathrm{\lambda }}_1=\frac{1}{\mathrm{R}} \end{gathered}$$

Step 2: Calculating relation between ${\mathrm{\lambda }}_1$ and ${\mathrm{\lambda }}_2$

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

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

$$\begin{gathered} \therefore \frac{1}{{\mathrm{\lambda }}_2}={\mathrm{R}}_{\mathrm{H}}\times {\mathrm{Z}}^2\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right) \\ \Rightarrow \frac{1}{{\mathrm{\lambda }}_2}={\mathrm{R}}_{\mathrm{H}}\times 4\left(\frac{1}{4}-\frac{1}{9}\right) \\ \Rightarrow \frac{1}{{\mathrm{\lambda }}_2}={\mathrm{R}}_{\mathrm{H}}\times \frac{5}{9} \\ \Rightarrow {\mathrm{\lambda }}_2={\mathrm{\lambda }}_1\times \frac{9}{5} \end{gathered}$$

Hence it is the correct option.

Therefore, option (A) $\frac{9{\mathrm{\lambda }}_1}{5}$ is the correct answer.