Question

The largest wavelength in the ultraviolet region of the hydrogen spectrum is $122\text{nm}$. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is

• A. $802\text{nm}$
• B. $823\text{nm}$
• C. $1882\text{nm}$
• D. $1648\text{nm}$

Answer 1

The correct option is B $823\text{nm}$

From Rydberg formula,

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

$\Rightarrow \lambda =\frac{1/R}{(\frac{1}{n_f^2}-\frac{1}{n_i^2})}\left[\text{For H- atom, Z=1}\right]$

The series in UV region is Lyman series, and we know that the longest wavelength will obtain corresponds to minimum energy which occurs in transition from $n_i=2$ to $n_f=1$.

$\therefore 122=\frac{1/R}{\frac{1}{1^2}-\frac{1}{2^2}}=\frac{4}{3R}.....\left(1\right)$

The smallest wavelength in the infrared region corresponds to maximum energy of Paschen series in transition from $n_i=\infty$ to $n_f=3$ will be

$\lambda =\frac{1/R}{\frac{1}{3^2}-\frac{1}{\infty }}=\frac{9}{R}........\left(2\right)$

From eqs. $\left(1\right)$ and $\left(2\right)$, we get

$\lambda =823.5\text{nm}$

Hence, option $\left(\text{B}\right)$ is correct.