The energy of the first (lowest) Balmer line | vedclass

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(D) The energy of a transition in a hydrogen-like atom is given by $\Delta E = 13.6 	imes Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) \ eV$.

Vedclass · Accepted Answer

$1.35x$

Vedclass · Answer

(D) The energy of a transition in a hydrogen-like atom is given by $\Delta E = 13.6 	imes Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) \ eV$. For the first Balmer line, the transition is from $n_2 = 3$ to $n_1 = 2$: $x = 13.6 	imes (1)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{9} ight) = 13.6 \left( \frac{5}{36} ight) \quad \dots(i)$ For the second Balmer line, the transition is from $n_2 = 4$ to $n_1 = 2$: $\Delta E = 13.6 	imes (1)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{16} ight) = 13.6 \left( \frac{3}{16} ight) \quad \dots(ii)$ Dividing equation $(ii)$ by equation $(i)$: $\frac{\Delta E}{x} = \frac{13.6 	imes (3/16)}{13.6 	imes (5/36)} = \frac{3}{16} 	imes \frac{36}{5} = \frac{3 	imes 9}{4 	imes 5} = \frac{27}{20} = 1.35$ Therefore, $\Delta E = 1.35x$.

The energy of the first (lowest) Balmer line of the $H$ atom is $x \ J$. The energy (in $J$) of the second Balmer line of the $H$ atom is:

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