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. When \[e\], \[m\] are the change and mass of electron in the \[n\]th orbit, \[h\] is Planck?s coustant and \[Z\] is the atomic number, then the energy of the electron in the electron in the \[n\]th orbit is \[E=-\frac{M\,{{Z}^{2}}{{e}^{4}}}{8\,{{\varepsilon }_{0}}{{h}^{2}}}.\frac{1}{{{n}^{2}}}\] 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, \[{{E}_{H}}={{E}_{n}}\] \[{{E}_{He}}=4{{E}_{n}}\]
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