VITEEE 2013 Paper

We have to compare wavelength of transition in the H-spectrum with the Balmer transition n = 4 to n = 2 of $He^{+}$ spectrum. Since $\lambda_H$ = $\lambda_{He^{+}}$ ∴ $R_HZ_H^2$ $\left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$ = $R_H Z_{He^{+}}^2$ $\left[ \frac{1}{2^2} - \frac{1}{4^2} \right]$ 1 × $\left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$ = 4 × $\left( \frac{1}{4} \times \frac{1}{16} \right)$ $\left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$ = 4 × $rac{4-1}{16}$ $\left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$ = $rac{3}{4}$ If $n_1$ = 1 , then $n_2$ = 2 , 3 , ... For first line $n_2$ = 2 , $n_1$ = 1 $\left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$ = $\frac{1}{1} - \frac{1}{4}$ = $rac{3}{4}$ Hence, trnasition $n_2$ = 2 to $n_1$ = 1 will give spectrum of the same wavelength as that of Balmer transition, $n_2$ = 4 to $n_1$ = 2 in $He^{+}$.