The wavelength of the first line of the Balme | vedclass

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Question

(A) For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$.
Using the Rydberg formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \fr

Vedclass · Accepted Answer

$6566$

Vedclass · Answer

(A) For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$.
Using the Rydberg formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.
$\frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R_H \left( \frac{1}{4} - \frac{1}{9} ight) = R_H \left( \frac{5}{36} ight)$.
$\lambda = \frac{36}{5R_H}$.
Given $R_H \approx 1.097 	imes 10^7 \ m^{-1} = 1.097 	imes 10^{-2} \ nm^{-1}$.
$\lambda = \frac{36}{5 	imes 1.097 	imes 10^{-2} \ nm^{-1}} \approx 656.3 \ nm = 6563 \ \mathring{A}$.
Rounding to the nearest provided option,the correct value is $6566 \ \mathring{A}$.

The wavelength of the first line of the Balmer series in the hydrogen spectrum is ................ $\mathring{A}$.

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