Question

Given below are three schematic graphs of potential energy $V (r)$ versus distance $r$ for three atomic particles : electron $(e^{-} )$, proton $(p^{+})$ 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

• A. $(1, n), (2, p^{+}), (3, e^{-} )$
• B. $(1, p^{+}), (2, e^{-}), (3, n)$
• C. $(1, e^{-}), (2, p^{+}), (3, n)$
• D. $(1, p^{+}), (2, n), (3, e^{- })$

Answer 1

Answer: (A)

For an electron and nucleus pair,
Potential energy$=\frac{K\left(-e\right)\left(+Ze\right)}{r}$
$=\frac{-KZe^{2}}{r}$
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 $(2).$