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(A) For an electron and nucleus pair,the potential energy is given by $V(r) = \frac{K(-e)(+Ze)}{r} = -\frac{KZe^2}{r}$.Since the potential energy is n

Vedclass · Accepted Answer

$(1, n), (2, p^{+}), (3, e^{-})$

Vedclass · Answer

(A) For an electron and nucleus pair,the potential energy is given by $V(r) = \frac{K(-e)(+Ze)}{r} = -\frac{KZe^2}{r}$.Since the potential energy is negative and approaches zero as $r$ increases,graph $(3)$ represents the electron.For a neutron,there is no electrostatic force outside the nucleus $(r > r_0)$. Thus,the potential energy is zero for $r > r_0$. Graph $(1)$ represents the neutron.For a proton,the electrostatic force is repulsive for $r > r_0$ because both the nucleus and the proton are positively charged. Thus,the potential energy is positive and decreases as $r$ increases. Graph $(2)$ represents the proton.Therefore,the correct pairing is $(1, n), (2, p^{+}), (3, e^{-})$,which corresponds to option $A$.

Similar Questions

The radius $R$ of a nucleus of mass number $A$ can be estimated by the formula $R = (1.3 \times 10^{-15}) A^{1/3} \; m$. It follows that the mass density of a nucleus is of the order of $(M_{\text{prot}} \cong M_{\text{neut}} = 1.67 \times 10^{-27} \; kg)$.

Which of the following statements is incorrect?

Two protons are kept at a separation of $10 \ nm$. Let $F_n$ and $F_e$ be the nuclear force and the electromagnetic force between them,respectively.

Why do stable nuclei never have more protons than neutrons?

$A$ reaction between a proton and $_8^{18}O$ that produces $_9^{18}F$ must also liberate:

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