Physics - Modern Physics Question with Solution | TestHub

Energy of photon $=\frac{\mathrm{hc}}{\lambda }=\frac{12500}{980}=12.75\mathrm{eV}$

Energy of electron in ${\mathrm{n}}^{\text{th }}$ orbit is given by

$\mathrm{En}=\frac{-13.6}{{\mathrm{n}}^2}\Rightarrow {\mathrm{E}}_{\mathrm{n}}-{\mathrm{E}}_1=-13.6\left[\frac{1}{{\mathrm{n}}^2}\frac{-1}{1^2}\right]$$\Rightarrow 12.75=13.6\left[\frac{1}{1^2}\frac{-1}{{\mathrm{n}}^2}\right]\Rightarrow \mathrm{n}=4$$\therefore$ Electron will excite to $\mathrm{n}=4$ We know that ' $R^{\prime }\propto n^2$$\therefore$ Radius of atom will be $16{\mathrm{a}}_0$