Ncert Exercises & Solutions

The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form of a Coulomb potential but with an effective charge e':

$F = \frac{1}{4 {πε}_{0}} \cdot \frac{e'^{2}}{r}$ $F = \frac{1}{4 {πε}_{0}} \cdot \frac{e'^{2}}{r}$

estimate the value of (e'/e) given that the binding energy of a deuteron is 2.2 MeV.

Hint: The binding energy depends on the charge and the mass of the nucleons.

Step 1: Find the binding energy of the H-atom.

The binding energy in H-atom,

E = \frac{{me}^{4}}{4 ε_{0}^{2} h^{2}} = 13.6 eV . . . . . . . . (i)

$E = \frac{{me}^{4}}{4 ε_{0}^{2} h^{2}} = 13.6 eV . . . . . . . . (i)$

If proton and neutron had charge e' each and were governed by the same electrostatic force, then in the above equation, we would need to replace electronic mass m with the reduced mass m' of proton-neutron and the electronic charge e by e'.

m' = \frac{M \times N}{M + N} = \frac{M}{2}

$m' = \frac{M \times N}{M + N} = \frac{M}{2}$

= \frac{1836 m}{2} = 918 m

$= \frac{1836 m}{2} = 918 m$

Here, M represents the mass of a neutron/proton.

Step 2: Find the binding energy of Deuteron.

∴ Binding energy = \frac{918 m {(e')}^{4}}{8 ε_{0}^{2} h^{2}} = 2.2 MeV . . . . . . . (ii)

$∴ Binding energy = \frac{918 m {(e')}^{4}}{8 ε_{0}^{2} h^{2}} = 2.2 MeV . . . . . . . (ii)$

Step 3: Find the ratio (e'/e).

Dividing Eqs. (ii) and (i), we get,

918 {(\frac{e'}{e})}^{4} = \frac{2.2 MeV}{13.6 eV} = \frac{2.2 \times 10^{6}}{13.6}

$918 {(\frac{e'}{e})}^{4} = \frac{2.2 MeV}{13.6 eV} = \frac{2.2 \times 10^{6}}{13.6}$

{(\frac{e'}{e})}^{4} = \frac{2.2 \times 10^{6}}{13.6} \times 918

${(\frac{e'}{e})}^{4} = \frac{2.2 \times 10^{6}}{13.6} \times 918$

\frac{e'}{e} = {(176.21)}^{\frac{1}{4}} = 3.64

$\frac{e'}{e} = {(176.21)}^{\frac{1}{4}} = 3.64$

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