Published by:
CGP EDU Academic Team
Published on: August 14, 2026
When an electron falls from (n + 2) state to (n) state in a He + ion the photon emitted has energy
6.172 × 10 –19 joules. What is the value of n.
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
For He + ion,
E n+2 – E n = 6.172 × 10 –19
∴ 13.6 2 
= 
∴ 13.6 
= 
= 0.966
Left side of above equation represents difference in energy of (n + 2) state and n state for Hydrogen atom and Right side of above equation represents difference in energy of 5 th state and 3 rd state for H atom.
∴ n = 3.
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