If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R = 0.1 Å and (ii) R = 10 Å .

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Answered by

Vishal Baghel | Contributor-Level 10

3 months ago

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

Explanation- in this type of situation centripetal force is balanced by electrostatic force

So mv²/r= -ke²/r²

According to bohr postuates

Mvr = h when n=1

On solving n $\frac{h^{2}}{m^{2} r^{2}} \frac{1}{r} = {k e}^{2} / r^{2}$

So after solving r= 0.51A⁰

So potential energy $\frac{{k e}^{2}}{r}$ = -27.2eV , KE= mv²/2

= $\frac{1}{2} m \frac{h^{2}}{r^{2}} = \frac{h}{2 m r^{2}} = + 13.6 e V$

But if R

If R>>r the electron moves inside the sphere with radius r’

Charge inside r’⁴=e $\frac{{r'}^{3}}{R^{3}}$

So r’ = $\frac{h^{2}}{m} \frac{k}{e^{2}} \frac{R^{3}}{{r'}^{3}}$

So r’⁴= 0.51A⁰R³

= 510(A⁰)⁴

This is a Long Answer Type Questions as classified in NCERT ExemplarExplanation- in this type of situation centripetal force is balanced by electrostatic forceSo mv2/r= -ke2/r2According to bohr postuatesMvr = h when n=1On solving n  <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <msup> <mrow> <mrow> <mi>k</mi> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>/</mo> <msup> <mrow> <mrow> <mi>r</mi> <mi> </mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> So after solving r= 0.51A0So potential energy  <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>k</mi> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </mfrac> </math> = -27.2eV , KE= mv2/2= <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mi>m</mi> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>m</mi> <msup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>=</mo> <mi> </mi> <mo>+</mo> <mn>13.6</mn> <mi>e</mi> <mi>V</mi> </math> But if RIf R>>r the electron moves inside the sphere with radius r’Charge inside r’4=e <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>r</mi> <mi>'</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> So r’ = <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> </mfrac> <mfrac> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>r</mi> <mi>'</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> So r’4= 0.51A0R3= 510(A0)4

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If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R = 0.1 Å and (ii) R = 10 Å .

1 Answer

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