JEE PYQ: Atomic Structure Question 41
Question 41 - 2019 (10 Jan Shift 2)
The ground state energy of hydrogen atom is $-13.6$ eV. The energy of second excited state of He$^+$ ion in eV is:
(1). ($-54.4$)
(2). ($-3.4$)
(3). ($-6.04$)
(4). ($-27.2$)
Show Answer
Answer: (3)
Solution
$E_3(\text{He}^+) = -13.6 \times \frac{4}{9} = -6.04$ eV.