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 $
