The potential energy(PE $_{n}$ ) of ion follows :

_{n}

α

_{π}

(\frac{n^{2}}{Z})

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α

_{π}

(\frac{Z^{2}}{n^{2}})

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_{n}

α

\frac{1}{m_{π}}

(\frac{n^{2}}{Z^{2}})

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_{n}

α

\frac{1}{m_{π}}

(\frac{Z^{2}}{n^{2}})

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Solution

The correct option is B PE $α$ m $_{π}$ $(\frac{Z^{2}}{n^{2}})$
Kinetic energy of pion $\Rightarrow K . E . = \frac{1}{2} m_{π} v^{2}$

Potential energy $\Rightarrow P . E . = - \frac{K Z e^{2}}{r}$

Hence, Total energy $\Rightarrow T . E = \frac{1}{2} m_{π} v^{2} - \frac{K Z e^{2}}{r}$

As $\frac{m_{π} v^{2}}{r} = \frac{K Z e^{2}}{r^{2}} \Rightarrow m_{π} v^{2} = \frac{K Z e^{2}}{r}$

$\Rightarrow P . E = m_{π} v^{2}$
As $v$ is directly proportional to $\frac{Z}{n}$ therefore, $P . E . α m_{π} \frac{Z^{2}}{n^{2}}$

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The potential energy(PEn) of ion follows :

The correct option is B PE α mπ (Z2n2)Kinetic energy of pion ⇒K.E.=12mπv2Potential energy ⇒P.E.=−KZe2rHence, Total energy ⇒T.E=12mπv2−KZe2rAs mπv2r=KZe2r2⇒mπv2=KZe2r⇒P.E=mπv2As v is directly proportional to Zn therefore, P.E. α mπZ2n2

The potential energy(PE $_{n}$ ) of ion follows :