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Chemistry-(general)

NEET Chemistry (general) MCQ Question

Type: MCQ-numerical-Medium-Class 11

Calculate the energy difference ΔE\Delta E in joules for an electron transition from n=5n = 5 to n=2n = 2 in a hydrogen atom.

A

4.58×10−19J4.58 \times 10^{-19} J

B

2.18×10−18J2.18 \times 10^{-18} J

C

1.55×10−19J1.55 \times 10^{-19} J

D

3.03×10−19J3.03 \times 10^{-19} J

Correct Answer

Option A

Detailed Explanation

In the hydrogen atom, the energy difference ΔE\Delta E for a transition from n=5n = 5 to n=2n = 2 is calculated as ΔE=2.18×10−18×(122−152)\Delta E = 2.18 \times 10^{-18} \times \left( \frac{1}{2^2} - \frac{1}{5^2} \right) J, resulting in 4.58×10−194.58 \times 10^{-19} J.

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