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 $
