Nuclear Physics And Radioactivity Question 74
Question: Excitation energy of a hydrogen like ion in its first excitation state is 40.8 eV. Energy needed to remove the electron from the ion in ground state is [KCET 2004]
Options:
A) 54.4 eV
B) 13.6 eV
C) 40.8 eV
D) 27.2 eV
Show Answer
Answer:
Correct Answer: A
Solution:
Excitation energy $ \Delta E=E_{2}-E_{1} $ $ =13.6\ Z^{2}[ \frac{1}{1^{2}}-\frac{1}{2^{2}} ] $
$ \Rightarrow 40.8=13.6\times \frac{3}{4}\times Z^{2}\Rightarrow Z=2. $ Now required energy to remove the electron from ground state $ =\frac{+13.6Z^{2}}{{{(1)}^{2}}}=13.6{{(Z)}^{2}}=54.4,eV. $
BETA
