Question 499: Atoms & Nuclei

Question

The ratio of maximum to minimum wavelength in Balmer series of hydrogen atom is

Accepted Answer

Maximum wavelength of Balmer series is given by
$$\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}ight)=\frac{5 \mathrm{R}}{36}\quad 	ext{.... (1)}$$
Maximum wavelength is given by
$$\frac{1}{\lambda^{\prime}}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{\infty}ight)=\frac{\mathrm{R}}{4} \quad 	ext{.... (2)}$$
Dividing Eq.(2) by Eq.(1)
$$\frac{\lambda}{\lambda^{\prime}}=\frac{9}{5}$$

Answer

36 : 5

Answer

3 : 4

Answer

9 : 5

Answer

5 : 9

Atoms & Nuclei

The ratio of maximum to minimum wavelength in Balmer series of hydrogen atom is

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