The wave number of the first emission line in | vedclass

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Question

(A) The wave number $\bar{
u}$ for a hydrogen-like species is given by the Rydberg formula: $\bar{
u} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^

Vedclass · Accepted Answer

$\frac{5}{36} R$

Vedclass · Answer

(A) The wave number $\bar{
u}$ for a hydrogen-like species is given by the Rydberg formula: $\bar{
u} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,$n_1 = 2$.The first emission line corresponds to the transition from $n_2 = 3$ to $n_1 = 2$.For the $H$ atom,$Z = 1$.Substituting these values: $\bar{
u} = R (1)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} ight)$.$\bar{
u} = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9 - 4}{36} ight) = \frac{5}{36} R$.

The wave number of the first emission line in the Balmer series of $H$ atom spectrum is ($R =$ Rydberg constant).

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