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$=\frac{K\left(-e\right)\left(+Ze\right)}{r}$
$=\frac{-KZ{e}^2}{r}$
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 $(2).$