JEE PYQ: Atomic Physics Question 49
Question 49 - 2019 (09 Apr Shift 2)
A He$^+$ ion is in its first excited state. Its ionization energy is:
(1) 48.36 eV
(2) 54.40 eV
(3) 13.60 eV
(4) 6.04 eV
Show Answer
Answer: (3)
Solution
$E_n = -13.6 \frac{Z^2}{n^2}$. For He$^+$, $E_2 = \frac{-13.6(2)^2}{2^2} = -13.60$ eV. Ionization energy $= 0 - E_2 = 13.60$ eV.