Q156 / 160 · TS EAMCET 2021 (06 Aug Shift 1) · Physics
A monochromatic radiation of wavelength $\lambda$ is incident on a hydrogen sample in ground state. The sample subsequently emits radiation of six different wavelengths, then the value of $\lambda$ is
[Use ch $=1242 \mathrm{eV}-\mathrm{m}$ ]
- A80 nm
- B85.5 nm
- Correct answerC97.4 nm
- D100.2 nm
Solution
Number of line emitted $=6, n_i=1, n_f=4$
$\begin{aligned}<br />\frac{n(n-1)}{2} & =6 \\<br />n^2-n-12 & =0 \\<br />(n-4)(n+3) & =0 \\<br />n=4, n & =-3 \quad \text{(Not possible)}<br />\end{aligned}$
Using, $\frac{1}{\lambda}=R \sqrt{\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)}$
$\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{4^2}\right)=R \frac{15}{16}$
where, $R=1.09 \times 10^7 \mathrm{~m}^{-1}$
$\lambda=\frac{16}{15} \times \frac{1}{1.09 \times 10^7}=97.4 \mathrm{~nm}$
$\begin{aligned}<br />\frac{n(n-1)}{2} & =6 \\<br />n^2-n-12 & =0 \\<br />(n-4)(n+3) & =0 \\<br />n=4, n & =-3 \quad \text{(Not possible)}<br />\end{aligned}$
Using, $\frac{1}{\lambda}=R \sqrt{\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)}$
$\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{4^2}\right)=R \frac{15}{16}$
where, $R=1.09 \times 10^7 \mathrm{~m}^{-1}$
$\lambda=\frac{16}{15} \times \frac{1}{1.09 \times 10^7}=97.4 \mathrm{~nm}$
Hence, the correct answer is option: 'C'.