Potential energy of electron present in $H e^{+}$ is:

\frac{e^{2}}{2 π ϵ_{0} r}

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\frac{3 e^{2}}{4 π ϵ_{0} r}

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\frac{- 2 e^{2}}{4 π ϵ_{0} r}

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\frac{- e^{2}}{4 π ϵ_{0} r}

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Solution

The correct option is C $\frac{- 2 e^{2}}{4 π ϵ_{0} r}$
$P . E = \frac{1}{4 π ϵ_{0}} \frac{(+ Z e) (- e)}{r}$
$= \frac{1}{4 π ϵ_{0}} \frac{(+ 2 e) (- e)}{r} = - \frac{e^{2}}{2 π ϵ_{0} r}$

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No/2 atoms of $X_{(g)}$ are converted into $X_{(g)^{+}}^{+}$ by energy $E_{1}$ , No/2 atoms of $X_{g}$ are converted into $X_{(g)}^{-}$ by energy $E_{2}$ . Hence ionization potential and electron affinity of $X_{(g)}$ per atom are

N_o/2 atoms of X(g) are converted into X⁺(g) by energy E₁. N_o/2 atoms of X(g) are converted into X^-(g) by energy E₂. Hence the ionisation potential and electron affinity of X(g) per atom are:

A. 2E₁/N_o., 2(E₂-E₁)/N_o

B. 2E₁/N_o.,. 2E₂/N_o

C. E₁-E₂/N_{o. ,.}2E₂/N_o

D. None is correct

Q. STATEMENT-1
For practical purposes, the earth is used as a reference at zero potential in electrical circuits.
and
STATEMENT-2
The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by