Atoms And Nuclei Question 426

Question: A hydrogen atom (ionisation potential 13.6 eV) makes a transition from third excited state to first excited state. The energy of the photon emitted in the process is

Options:

A) 1.89 eV

B) 2.55 eV

C) 12.09 eV

D) 12.75 eV

Show Answer

Answer:

Correct Answer: B

Solution:

  • Energy released $ =13.6[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(4)}^{2}}} ]=2.55,eV $


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