A) \[20eV\]
B) \[79eV\]
C) \[109eV\]
D) \[34eV\]
Correct Answer: B
Solution :
Let \[{{E}_{1}}=\]energy required to remove \[{{e}^{-}}\] from singly ionised Helium atom \[=\frac{+13.6{{z}^{2}}}{{{n}^{2}}}\] \[=\frac{(13.6){{(2)}^{2}}}{{{(1)}^{2}}}\] \[=54.4eV\] \[{{E}_{2}}=\] Energy required to remove an \[{{e}^{-}}\] from He ?atom Given \[{{E}_{1}}=2.2{{E}_{2}}\] \[\therefore {{E}_{2}}=\frac{{{E}_{1}}}{2.2}=\frac{54.4}{2.2}=24.7eV\] \[\therefore \] Energy required to remove both \[{{e}^{-}}'s\] from He-atom \[=24.7+54.4\] \[=79.1eV\]You need to login to perform this action.
You will be redirected in
3 sec