Question

If energy required to remove one of the two electron from $\text{He}$ atom is $29.5eV,$ then what is the value of energy required to convert a helium atom into $\alpha -$particle?

• A. $54.4eV$
• B. $83.9eV$
• C. $29.5eV$
• D. $24.9eV$

Answer 1

The correct option is B $83.9eV$

From a He-atom, containing $2e^{-}$ m ground state, one electron is

removed by proving 29.5 ev of energy, and remain $H{e}^{+}$

$He\stackrel{29.sev}{\to }H{e}^{+}+e^{-}$

$H{e}^{+}$ (single $e^{-}$ species ) because uke Bohr H-atom having ground

state energy,

$E_1=-13.6\times \frac{z^2}{n^2}$ , for $z=2\left(He\right)2n=1$

$=-13.6\times \frac{z^2}{12}=-54.4ev$

So, an addition 54.4ev should be given to convent $H{e}^{+}$ to

$\alpha -$ partial (which is basically He nuclei, or $H{e}^{++}$).

$H{e}^{+}\stackrel{54.4ev}{\to }H{e}^{++}+e^{-}$

Net energy required $=29.5+54.4=83.9ev$