JEE PYQ: Atomic Physics Question 59
Question 59 - 2019 (11 Jan Shift 1)
A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980 A. The radius of the atom in the excited state, in terms of Bohr radius $a_0$, will be:
($hc = 12500$ eV-A)
(1) $25a_0$
(2) $9a_0$
(3) $16a_0$
(4) $4a_0$
Show Answer
Answer: (3)
Solution
Energy of photon $= \frac{hc}{\lambda} = \frac{12500}{980} = 12.75$ eV. Energy of electron in $n^{th}$ orbit: $E_n = \frac{-13.6}{n^2}$. $E_n - E_1 = -13.6\left[\frac{1}{n^2} - 1\right]$. $12.75 = 13.6\left[1 - \frac{1}{n^2}\right] \Rightarrow n = 4$. Electron will excite to $n = 4$. Radius $\propto n^2$. Radius $= 16a_0$.
BETA
