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 a deuteron as given in the form a coulomb potential but with an effective charge e’.

F= ke’²/r estimate the value of e’/e given that the the binding energy of a duetron is 2.2MeV.

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alok kumar singh | Contributor-Level 10

4 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation-E= $\frac{m e^{4}}{π ε_{0}^{2} h^{2}} = 13.6 e V$

If proton and neutron had charge e each and were governed by the same electrostatic force, then in the above equation we would need relace m

So m’ = $\frac{M \times N}{M + N}$ =M/2=918m

Hence binding energy= 918me’⁴/8 $ε_{0}^{2} h^{2} = 2.2 M e V$

Dividing eqn

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

$\frac{e'}{e}$ = 3.64

This is a Long Answer Type Questions as classified in NCERT ExemplarExplanation-E= <math><mfrac><mrow><mrow><mi>m</mi><msup><mrow><mrow><mi>e</mi></mrow></mrow><mrow><mrow><mn>4</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mi>π</mi><msubsup><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msubsup><msup><mrow><mrow><mi>h</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac><mo>=</mo><mn>13.6</mn><mi>e</mi><mi>V</mi></math>If proton and neutron had charge e each and were governed by the same electrostatic force, then in the above equation we would need relace mSo m’ = <math><mfrac><mrow><mrow><mi>M</mi><mo>×</mo><mi>N</mi></mrow></mrow><mrow><mrow><mi>M</mi><mo>+</mo><mi>N</mi></mrow></mrow></mfrac></math> =M/2=918mHence binding energy= 918me’4/8 <math> <msubsup> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <msup> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>=</mo> <mn>2.2</mn> <mi>M</mi> <mi>e</mi> <mi>V</mi> </math> Dividing eqn918 ( <math> <mfrac> <mrow> <mrow> <mi>e</mi> <mi>'</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </mfrac> <mo>)</mo> </math> 4= <math> <mfrac> <mrow> <mrow> <mn>2.2</mn> <mi>M</mi> <mi>e</mi> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>13.6</mn> <mi>e</mi> <mi>V</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>2.2</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>13.6</mn> </mrow> </mrow> </mfrac> </math> <math><mfrac><mrow><mrow><mi>e</mi><mi>'</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></mfrac></math>= 3.64

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