An electron jumps from the 5^{th} orbit to th | vedclass

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(C) The frequency $
u$ of the emitted radiation is given by the formula: $
u = RC \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$Here,$R = 10^7 \

Vedclass · Accepted Answer

$6.75 	imes 10^{13} \ Hz$

Vedclass · Answer

(C) The frequency $
u$ of the emitted radiation is given by the formula: $
u = RC \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$Here,$R = 10^7 \ m^{-1}$,$C = 3 	imes 10^8 \ m/s$,$n_1 = 4$,and $n_2 = 5$.Substituting the values:$
u = (10^7) 	imes (3 	imes 10^8) \left[ \frac{1}{4^2} - \frac{1}{5^2} ight]$$
u = 3 	imes 10^{15} \left[ \frac{1}{16} - \frac{1}{25} ight]$$
u = 3 	imes 10^{15} \left[ \frac{25 - 16}{400} ight]$$
u = 3 	imes 10^{15} \left[ \frac{9}{400} ight]$$
u = \frac{27}{400} 	imes 10^{15} = 0.0675 	imes 10^{15} \ Hz = 6.75 	imes 10^{13} \ Hz$.

An electron jumps from the $5^{th}$ orbit to the $4^{th}$ orbit of a hydrogen atom. Taking the Rydberg constant as $10^7 \ m^{-1}$,what will be the frequency of the radiation emitted?

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