Atoms And Nuclei Question 132

Question: Energy E of a hydrogen atom with principal quantum number n is given by $ E=\frac{-13.6}{n^{2}}eV $ . The energy of a photon ejected when the electron jumps from $ n=3 $ state to $ n=2 $ state of hydrogen is approximately [CBSE PMT 2004]

Options:

A) 1.5 eV

B) 0.85 eV

C) 3.4 eV

D) 1.9 eV

Show Answer

Answer:

Correct Answer: D

Solution:

$ {E_{3\to 2}}=-3.4-(-1.51)=-1.89,eV $ Þ $ |{E_{3\to 2}}|\ \approx 1.9,eV $



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