Practice free →
HomeNEET UG › Atomic Structure › When a hydrogen atom makes a transition from n =…

When a hydrogen atom makes a transition from n = 2 to n = 1, the energy of the emitted photon is approximately:

A10.2 eV
B3.4 eV
C1.51 eV
D13.6 eV
Answer & Solution
Correct answer: A. 10.2 eV
ΔE = E_2 − E_1 = (−3.4) − (−13.6) = 10.2 eV.
Solve this in the app — NEET UG practice & 24k+ MCQs →
Related questions