The Bohr model for the H-atom relies on the Coulomb ’s law of electrostatics. Coulomb’s law has not directly been verified for very
short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge +q₁, – q₂modified to

F= $\frac{{k q}_{1} q_{2}}{r^{2}}$ r>R₀

= $\frac{{k q}_{1} q_{2}}{R_{0}^{2}}$ ( ${\frac{R_{o}}{r})}^{ε}$ r0

Calculate in such vase the ground state energy of a H atom , if E= 0.1 R₀= 1A⁰

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Vishal Baghel | Contributor-Level 10

4 months ago

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

Explanation- when r0=1A^o

Let $ε = 2 + δ$

F= $\frac{{k q}_{1} q_{2}}{r^{2 + δ}} R_{0}^{δ}$

Where $\frac{{k q}_{1} q_{2}}{r^{2}}$ = ( ${1.6 \times 10^{- 19})}^{2} \times 9 \times 10^{9} = 23.04 \times 10^{- 29} N / m 2$

Let p= $23.04 \times 10^{- 29} N / m 2$

Electrostatic force is balanced by centripetal force

Mv²/r or v²= $\frac{p R_{0}^{δ}}{m r^{1 + δ}}$

This is a Long Answer Type Questions as classified in NCERT ExemplarExplanation- when r<r0=1Ao</rLet <math> <mi>ε</mi> <mo>=</mo> <mn>2</mn> <mo>+</mo> <mi>δ</mi> </math> F= <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>k</mi> <mi>q</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>q</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mo>+</mo> <mi>δ</mi> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <msubsup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mi>δ</mi> </mrow> </mrow> </msubsup> </math> Where <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>k</mi> <mi>q</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>q</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = ( <math> <msup> <mrow> <mrow> <mn>1.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>9</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>9</mn> </mrow> </mrow> </msup> <mo>=</mo> <mn>23.04</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>29</mn> </mrow> </mrow> </msup> <mi>N</mi> <mo>/</mo> <mi>m</mi> <mn>2</mn> </math> Let p= <math> <mn>23.04</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>29</mn> </mrow> </mrow> </msup> <mi>N</mi> <mo>/</mo> <mi>m</mi> <mn>2</mn> </math> Electrostatic force is balanced by centripetal forceMv2/r or v2= <math> <mfrac> <mrow> <mrow> <mi>p</mi> <msubsup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mi>δ</mi> </mrow> </mrow> </msubsup> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> <msup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>+</mo> <mi>δ</mi> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>

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