An energy of $$24.6\,eV$$ is required to remove one of the electrons from a neutral helium atom. The energy in $$\left( {eV} \right)$$ required to remove both the electrons from a neutral helium atom is
A.
38.2
B.
49.2
C.
51.8
D.
79.0
Answer :
79.0
Solution :
When one $${e^ - }$$ is removed from neutral helium atom, it becomes a one $${e^ - }$$ species.
For one $${e^ - }$$ species we know
$${E_n} = \frac{{ - 13.6{Z^2}}}{{{n^2}}}eV/atom$$
For helium ion, $$Z = 2$$ and for first orbit $$n = 1.$$
$$\therefore {E_1} = \frac{{ - 13.6}}{{{{\left( 1 \right)}^2}}} \times {2^2} = - 54.4eV$$
$$\therefore $$ Energy required to remove this $${e^ - } = + 54.4\,eV$$
$$\therefore $$ Total energy required $$ = 54.4 + 24.6 = 79eV$$
Releted MCQ Question on Modern Physics >> Atoms And Nuclei
Releted Question 1
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