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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.


Learning Progress: Step 41 of 45 in this series