A) \[4{{E}_{n}}\]
B) \[{{E}_{n}}/4\]
C) \[2{{E}_{n}}\]
D) \[{{E}_{n}}/2\]
Correct Answer: A
Solution :
Key Idea: Atomic number of helium is twice that of hydrogen. The energy of the electron in the nth orbit is \[E=E\propto {{Z}^{2}}.\frac{1}{{{n}^{2}}}\] Given, \[{{Z}_{H}}=1,\,{{Z}_{He}}=2\] \[\therefore \] \[\frac{{{E}_{H}}}{{{E}_{He}}}=\frac{Z_{H}^{2}}{Z_{He}^{2}}=\frac{1}{4}\] \[\Rightarrow \] \[{{E}_{He}}=4{{E}_{H}}\] Given, \[{{E}_{H}}={{E}_{n}}\] \[\therefore \]\[{{E}_{He}}=4{{E}_{n}}\]
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