A) 54.4 eV
B) 13.6 eV
C) 40.8 eV
D) 27.2 eV
Correct Answer: A
Solution :
Excitation energy \[\Delta E={{E}_{2}}-{{E}_{1}}\]\[=13.6\ {{Z}^{2}}\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right]\] \[\Rightarrow 40.8=13.6\times \frac{3}{4}\times {{Z}^{2}}\Rightarrow Z=2.\] Now required energy to remove the electron from ground state \[=\frac{+13.6{{Z}^{2}}}{{{(1)}^{2}}}=13.6{{(Z)}^{2}}=54.4\,eV.\]You need to login to perform this action.
You will be redirected in
3 sec