CUET Preparation Today
CUET
Physics
Atoms
a
b
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d
$\lambda_1 = \frac{hc}{-E+2E} = \frac{hc}{E}$
$\lambda_2 = \frac{hc}{-E+4E/3} = \frac{3hc}{E}$
$\frac{\lambda_1}{\lambda_2} = \frac{1}{3}$
