Question 455: Atoms & Nuclei

Question

In hydrogen atom, ratio of the shortest wavelength in the Balmer series to that in the Paschen series is

Accepted Answer

The shortest wavelength in Balmer series is given by
$$\frac{1}{\lambda_1}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{\infty}ight)=\frac{\mathrm{R}}{4}\quad	ext{.... (i)}$$
The shortest wavelength in the Paschen series given by
$$\frac{1}{\lambda_2}=\mathrm{R}\left(\frac{1}{3^2}-\frac{1}{\infty}ight)=\frac{R}{9}\quad	ext{.... (ii)}$$
Dividing equation (ii) by equation (i),
$$\frac{\lambda_1}{\lambda_2}=\frac{\mathrm{R}}{9} 	imes \frac{4}{\mathrm{R}}=\frac{4}{9}$$

Answer

$9: 4$

Answer

$3: 1$

Answer

$4: 9$

Answer

$1: 3$

Atoms & Nuclei

In hydrogen atom, ratio of the shortest wavelength in the Balmer series to that in the Paschen series is

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