A) \[\frac{{{E}_{n}}}{2}\]
B) \[2{{E}_{n}}\]
C) \[4{{E}_{n}}\]
D) \[\frac{{{E}_{n}}}{4}\]
Correct Answer: C
Solution :
The energy of election of charge \[e\], mass \[m\], in an orbit is the sum of kinetic and potential energies. The total energy in the \[nth\] orbit is given by \[E=-\frac{m{{Z}^{2}}{{e}^{4}}}{8\varepsilon _{0}^{2}{{h}^{2}}}\left( \frac{1}{{{n}^{2}}} \right)\] Where Z is atomic number. For helium \[Z=2\] \[\therefore \] \[E\propto {{Z}^{2}}\propto {{\left( 2 \right)}^{2}}\propto 4\] \[\Rightarrow \] \[E=4\,{{E}_{n}}\]
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