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(B) The process of converting a neutral $He$ atom into an $\alpha$-particle $(He^{2+})$ involves two steps:$1$. Removing the first electron from the n

Vedclass · Accepted Answer

$83.9$

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(B) The process of converting a neutral $He$ atom into an $\alpha$-particle $(He^{2+})$ involves two steps:$1$. Removing the first electron from the neutral $He$ atom: $E_1 = 29.5 \,eV$ (given).$2$. Removing the second electron from the $He^+$ ion: The energy required to remove an electron from a hydrogen-like ion is given by $E_n = 13.6 	imes Z^2 / n^2 \,eV$. For $He^+$,$Z = 2$ and $n = 1$,so $E_2 = 13.6 	imes 2^2 = 13.6 	imes 4 = 54.4 \,eV$.Total energy required $E_{	ext{total}} = E_1 + E_2 = 29.5 \,eV + 54.4 \,eV = 83.9 \,eV$.

If the energy required to remove one of the two electrons from a $He$ atom is $29.5 \,eV$,then what is the total energy required to convert a helium atom into an $\alpha$-particle (i.e.,$He^{2+}$ ion)?

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