Q51 / 66 · JEE Advanced 2007 (Paper 1) · Physics
The largest wavelength in the ultraviolet region of the hydrogen spectrum is $122 \mathrm{~nm}$. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is
- A$802 \mathrm{~nm}$
- Correct answerB$823 \mathrm{~nm}$
- C$1882 \mathrm{~nm}$
- D$1648 \mathrm{~nm}$
Solution
The series in $U-V$ region is Lymen series. Longest wavelength corresponds to minimum energy which occurs in transition from $n=2$ to $n=1$.
$$
\therefore \quad 122=\frac{\frac{1}{R}}{\left(\frac{1}{1^2}-\frac{1}{2^2}\right)}
$$
The smallest wavelength in the infrared region corresponds to maximum energy of Paschen series.
$$
\lambda=\frac{\frac{1}{R}}{\left(\frac{1}{3^2}-\frac{1}{\infty}\right)}
$$
Solving Eqs. (i) and (ii), we get
$\therefore$ Correct option is (b).
$$
\lambda=823.5 \mathrm{~nm}
$$
Hence, the correct answer is option: 'B'.