Ncert Exercises & Solutions

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, -q₂ is modified to

$| F | = \frac{q_{1} q_{2}}{(4 {πε}_{0})} \frac{1}{r^{2}}, r \geq R_{0}$ $|F| = \frac{q_{1} q_{2}}{(4 {πε}_{0})} \frac{1}{r^{2}}, r \geq R_{0}$
$= \frac{q_{1} q_{2}}{4 {πε}_{0}} \frac{1}{R_{0}^{2}} {(\frac{R_{0}}{r})}^{ε}, r \leq R_{0}$ $= \frac{q_{1} q_{2}}{4 {πε}_{0}} \frac{1}{R_{0}^{2}} {(\frac{R_{0}}{r})}^{ε}, r \leq R_{0}$

Calculate in such a case, the ground state energy of a H-atom, if $ε$ $ε$ = 0.1, $R_{0} = 1 Å .$ $R_{0} = 1 Å .$

Hint: The electrostatic force of attraction between the positively charged nucleus and negatively charged electrons provides the necessary centripetal force.

Step 1: Find the radius of the orbit of the electron in the ground state.

Considering the case, when

r \leq R_{0} = 1 Å

$r \leq R_{0} = 1 Å$

Let ε = 2 + δ

$Let ε = 2 + δ$

F = \frac{q_{1} q_{2}}{4 {πε}_{0}} \frac{R_{0}^{δ}}{r^{2 + δ}}

$F = \frac{q_{1} q_{2}}{4 {πε}_{0}} \frac{R_{0}^{δ}}{r^{2 + δ}}$

where, \land = \frac{q_{1} q_{2}}{4 {πε}_{0}} = {(1.6 \times 10^{- 19})}^{2} \times 9 \times 10^{9}

$where, \land = \frac{q_{1} q_{2}}{4 {πε}_{0}} = {(1.6 \times 10^{- 19})}^{2} \times 9 \times 10^{9}$

≃ 23.04 \times 10^{- 29} N m^{2}

$≃ 23.04 \times 10^{- 29} N m^{2}$

The electrostatic force of attraction between the positively charged nucleus and negatively charged electrons (Coulombian force) provides the necessary centripetal force.

F = \frac{{mv}^{2}}{r} = \frac{\land R_{0}^{δ}}{r^{2 + δ}} or v^{2} = \frac{\land R_{0}^{δ}}{{mr}^{1 + δ}}

$F = \frac{{mv}^{2}}{r} = \frac{\land R_{0}^{δ}}{r^{2 + δ}} or v^{2} = \frac{\land R_{0}^{δ}}{{mr}^{1 + δ}}$

mvr = nħ \Rightarrow r = \frac{nħ}{mv} = \frac{nħ}{m} {[\frac{m}{\land R_{0}^{δ}}]}^{1 / 2} r^{1 / 2 + δ / 2} [Applying Bohr' s second postulates]

$mvr = nħ \Rightarrow r = \frac{nħ}{mv} = \frac{nħ}{m} {[\frac{m}{\land R_{0}^{δ}}]}^{1 / 2} r^{1 / 2 + δ / 2} [Applying Bohr' s second postulates]$

Solving this for r, we get

$Solving this for r, we get$

r_{n} = {[\frac{n^{2} ħ^{2}}{m \land R_{0}^{δ}}]}^{\frac{1}{1 - δ}} where, r_{n} is radius of nth orbit of electron .

$r_{n} = {[\frac{n^{2} ħ^{2}}{m \land R_{0}^{δ}}]}^{\frac{1}{1 - δ}} where, r_{n} is radius of nth orbit of electron .$

For n = 1 and substituting the values of constant, we get

$For n = 1 and substituting the values of constant, we get$

r_{1} = {[\frac{ħ^{2}}{m \land R_{0}^{δ}}]}^{\frac{1}{1 - δ}}

$r_{1} = {[\frac{ħ^{2}}{m \land R_{0}^{δ}}]}^{\frac{1}{1 - δ}}$

r_{1} = {[\frac{1 . 05^{2} \times 10^{- 68}}{9.1 \times 10^{- 31} \times 2.3 \times 10^{- 28} \times 10^{+ 19}}]}^{\frac{1}{2.9}}

$r_{1} = {[\frac{1 . 05^{2} \times 10^{- 68}}{9.1 \times 10^{- 31} \times 2.3 \times 10^{- 28} \times 10^{+ 19}}]}^{\frac{1}{2.9}}$

= 8 \times 10^{- 11}

$= 8 \times 10^{- 11}$

= 0.08 nm (< 0.1 nm)

$= 0.08 nm (< 0.1 nm)$

This is the radius of orbit of the electron in the ground state of the hydrogen atom.

Step 2: Find the kinetic energy and the potential energy of the electron in the ground state.

v_{n} = \frac{nħ}{{mr}_{n}} = n ħ {(\frac{m \land R_{0}^{δ}}{n^{2} ħ^{2}})}^{\frac{1}{1 - δ}}

$v_{n} = \frac{nħ}{{mr}_{n}} = n ħ {(\frac{m \land R_{0}^{δ}}{n^{2} ħ^{2}})}^{\frac{1}{1 - δ}}$

For n = 1, v_{1} = \frac{ħ}{{mr}_{1}} = 1.44 \times 10^{6} m / s [This is the speed of electron in ground state]

$For n = 1, v_{1} = \frac{ħ}{{mr}_{1}} = 1.44 \times 10^{6} m / s [This is the speed of electron in ground state]$

$KE = \frac{1}{2} {mv}_{1}^{2} = 9.43 \times 10^{- 19} J = 5.9 eV [This is the KE of electron in ground state]$

$PE till R_{0} = - \frac{\land}{R_{0}} [This is the PE of electron in ground state at r = R_{0}]$

$PE from R_{0} to r = + \land R_{0}^{δ} \int_{R_{0}}^{r} \frac{dr}{r^{2 + δ}} = + \frac{\land R_{0}^{δ}}{- 1 - δ} {[\frac{1}{r^{1 + δ}}]}_{R_{0}}^{r} [This is the PE of electron in ground state at R_{0} to r]$

$= - \frac{\land R_{0}^{δ}}{1 + δ} [\frac{1}{r^{1 + δ}} - \frac{1}{R_{0}^{1 + δ}}] = - \frac{\land}{1 + δ} [\frac{R_{0}^{δ}}{r^{1 + δ}} - \frac{1}{R_{0}}]$

$PE = - \frac{\land}{1 + δ} [\frac{R_{0}^{δ}}{r^{1 + δ}} - \frac{1}{R_{0}} + \frac{1 + δ}{R_{0}}]$

$PE = - \frac{\land}{- 0.9} [\frac{R_{0}^{- 1.9}}{r^{- 0.9}} - \frac{1.9}{R_{0}}]$

$= \frac{2.3}{0.9} \times 10^{- 18} [{(0.8)}^{0.9} - 1.9] J = - 17.3 eV$

Total energy is (-17.3 + 5.9)=-11.4 eV

This is the required TE of an electron in the ground state.

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