CUET Preparation Today
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$\text{Ionization Energy } E_n= \frac{kz^2R}{n^2}$ $\text{It is given that } kRz^2(1 - \frac{1}{\infty^2}) = kRZ^2= 9R$ $ E_3 - E_1 = kRz^2 (1 - \frac{1}{3^2}) = \frac{8kz^2R}{9} = 8R =\frac{hc}{\lambda}$ $\Rightarrow \lambda = 11.4nm$ |
