- Home
- Standard 12
- Physics
Given below are three schematic graphs of potential energy $V(r)$ versus distance $r$ for three atomic particles : electron $\left(e^{-}\right)$, proton $\left(p^{+}\right)$and neutron $(n)$, in the presence of a nucleus at the origin $O$. The radius of the nucleus is $r_0$. The scale on the $V$-axis may not be the same for all figures. The correct pairing of each graph with the corresponding atomic particle is
$(1, n),\left(2, p^{+}\right),\left(3, e^{-}\right)$
$\left(1, p^{+}\right),\left(2, e^{-}\right),(3, n)$
$\left(1, e^{-}\right),\left(2, p^{+}\right),(3, n)$
$\left(1, p^{+}\right),(2, n),\left(3, e^{-}\right)$
Solution
(a)
For an electron and nucleus pair, Potential energy $=\frac{K(-e)(+Z e)}{r}$
$=\frac{-K Z e^2}{r}$
So, potential energy of electron is negative and it tends to zero as separation $r$ increases.
Hence, correct variation of potential energy with $r$ is as shown in graph $(3)$. For a neutron, force outside nucleus is zero.
Hence, potential energy of neutron is zero as $r > r_0$.
So, correct variation of potential energy with $r$ for a neutron is as shown in graph $(1)$.
For a proton, as $r > r_0$, force is repulsive. Hence, potential energy is positive.
So, correct variation of potential energy with $r$ is as shown in graph $(2)$.
