હાઇડ્રોજન પરમાણુના ક્યા સંક્રમણની તરંગલંબાઇ $He^+$ ના $n = 4$ થી $n = 2$ સંક્રમણની તરંગલંબાઇને સમાન થશે ?

Question

Medium

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Solution

b
$H$ $He ^{+}$

$\lambda_1=\lambda_2$

$\therefore \frac{1}{\lambda_1}=\frac{1}{\lambda_2}$

$\therefore\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \times Z_1^2=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \times Z_2^2$

$\therefore\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \times 1^2=\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \times 2^2$

$\therefore \frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{1}-\frac{1}{4}=\frac{1}{1}-\frac{1}{2^2}$

$\therefore n_1=1, n_2=2$

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