Question 495: Atoms & Nuclei

Question

The wave number of the last line of the Balmer series in hydrogen spectrum will be
(Rydberg's constant $$=10^7 \mathrm{~m}^{-1}$$ )

Accepted Answer

Wave number $$=\frac{1}{\lambda}=$$ Receiprocal of wavelength
For the last line of the Balmer series, $$n=\infty$$ and the transition is from $$n=\infty$$ to $$n=2$$
$$	herefore$$ Wave number $$\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{\infty}ight]=\frac{\mathrm{R}}{4}$$ and $$\mathrm{R}=10^7 / \mathrm{m}$$
$$	herefore \bar{v}=\frac{R}{4}=\frac{10^7}{4}=\frac{10 	imes 10^6}{4}=2.5 	imes 10^6 \mathrm{~m} / \mathrm{s}$$

Answer

$$250 \mathrm{~m}^{-1}$$

Answer

$$2.5 	imes 10^6 \mathrm{~m}^{-1}$$

Answer

$$0.25 	imes 10^9 \mathrm{~m}^{-1}$$

Answer

$$2.5 	imes 10^5 \mathrm{~m}^{-1}$$

Atoms & Nuclei

The wave number of the last line of the Balmer series in hydrogen spectrum will be (Rydberg's constant $=10^7 \mathrm{~m}^{-1}$ )

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