Question 392: Atoms & Nuclei

Question

In the hydrogen spectrum, $$\lambda$$ be the wavelength of first transition line of Lyman series. The wavelength difference will be "a$$\lambda$$'' between the wavelength of $$3^{	ext {rd }}$$ transition line of Paschen series and that of $$2^{	ext {nd }}$$ transition line of Balmer series where $$\mathrm{a}=$$ ___________.

Accepted Answer

$${1 \over \lambda } = {R_H}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} ight)$$
$${1 \over {{\lambda _3}}} = {R_H}\left( {{1 \over {{3^2}}} - {1 \over {{6^2}}}} ight)$$
$${1 \over {{\lambda _2}}} = {R_H}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} ight)$$
$$	herefore$$ $${\lambda _3} - {\lambda _2} = a\lambda $$
$$a = 5$$

Atoms & Nuclei

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