Question 414: Atoms & Nuclei

Question

The ratio of minimum wavelengths of Lyman and Balmer series will be

Accepted Answer

Wavelength, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}^2}-\frac{1}{\mathrm{~m}^2}ight)$
For shortest wavelength in Lyman series:
$$\begin{aligned}
& n=1, m=\infty \
& \frac{1}{\lambda_L}=R\left(\frac{1}{1^2}-\frac{1}{\infty^2}ight)=R\quad	ext{.... (i)}
\end{aligned}$$
For shortest wavelength in Balmer series:
$$\begin{aligned}
& \mathrm{n}=2, \mathrm{~m}=\infty \
& \frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{\infty^2}ight)=\frac{\mathrm{R}}{4}\quad	ext{.... (ii)}\end{aligned}$$
Dividing equation (ii) by equation (i),
$$\frac{\lambda_1}{\lambda_B}=\frac{\mathrm{R} / 4}{\mathrm{R}}=\frac{1}{4}=0.25$$

Answer

1.25

Answer

0.25

Answer

5

Answer

10

Atoms & Nuclei

The ratio of minimum wavelengths of Lyman and Balmer series will be

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