Ncert Exercises & Solutions

The inverse square law in electrostatic is $|F| = \frac{e^{2}}{(4 {πε}_{0}) r^{2}}$ for the force between an electron and a proton. The $(\frac{1}{r})$ dependence of $|F|$ can be understood in quantum theory as being due to the fact that the particle of light (photon) is massless. If photons had a mass m_p, the force would be modified to $|F| = \frac{e^{2}}{(4 {πε}_{0}) r} [\frac{1}{r^{2}} + \frac{λ}{r}] \cdot \exp (- λ r)$ where $λ = \frac{m_{p} c}{ħ}$ and $ħ = \frac{h}{2 π}$ . Estimate the change in the ground state energy of a H-atom if m_p were 10^-6 times the mass of an electron.

Hint: The electrostatic force provides the required centripetal force.

Step 1: Find the wavelength.

For

m_{p} = 10^{- 6} times

the mass of an electron, the energy associated with it is given by-

m_{p} c^{2} = 10^{- 6} \times electron mass \times c^{2}

≃ 10^{- 6} \times 0.5 MeV

≃ 10^{- 6} \times 0.5 \times 1.6 \times 10^{- 23}

≃ 0.8 \times 10^{- 19} J

The wavelength associated with it is given by:

\frac{ħ}{m_{p} c} = \frac{ħ c}{m_{p} c^{2}} = \frac{10^{- 34} \times 3 \times 10^{8}}{0.8 \times 10^{- 19}}

\approx 4 \times 10^{- 7} m > > Bohr radius

Step 2: Find the potential energy.

|F| = \frac{e^{2}}{4 {πε}_{0}} [\frac{1}{r^{2}} + \frac{λ}{r}] \exp (- λ r)

w here, λ^{- 1} = \frac{ħ}{m_{p} c} \approx 4 \times 10^{- 7} m > > r_{B}

∴ λ < < \frac{1}{r_{B}} i . e ., λ r_{B} < < 1

U (r) = - \frac{e^{2}}{4 {πε}_{0}} \cdot \frac{\exp (- λ r)}{r}

m v r = ħ ∴ v = \frac{ħ}{m r}

A l s o \frac{m v^{2}}{r} = \approx (\frac{e^{2}}{4 {πε}_{0}}) [\frac{1}{r^{2}} + \frac{λ}{r}]

∴ \frac{ħ^{2}}{m r^{3}} = (\frac{e^{2}}{4 {πε}_{0}}) [\frac{1}{r^{2}} + \frac{λ}{r}]

∴ \frac{ħ^{2}}{m} = (\frac{e^{2}}{4 {πε}_{0}}) [r + π r^{2}]

if λ = 0; r = r_{B} = \frac{ħ}{m} \cdot \frac{4 {πε}_{0}}{e^{2}}

\frac{ħ^{2}}{m} = \frac{e^{2}}{4 {πε}_{0}} \cdot r

Since, λ^{- 1} > > r, put r = r_{B} + δ

∴ r_{B} = r_{B} + δ + λ (r_{B}^{2} + δ^{2} + 2 {δr}_{B}); neglect δ^{2}

or 0 = {λr}_{B}^{2} + δ (1 + 2 {λr}_{B})

δ = \frac{- {λr}_{B}^{2}}{1 + 2 {λr}_{B}} \approx {λr}_{B}^{2} (1 - 2 {λr}_{B}) = - {λr}_{B}^{2}

Since, λ r_{B} < < 1

∴ V (r) = - \frac{e^{2}}{4 {πε}_{0}} \cdot \frac{e x p (- λδ - λ r_{B})}{r_{B} + δ}

∴ V (r) = - \frac{e^{2}}{4 {πε}_{0}} \frac{1}{r_{B}} [(1 - \frac{δ}{r_{B}}) \cdot (1 - λ r_{B})]

≅ (- 27.2 eV) remains unchanged

Step 3 : Find the kinetic energy .

KE = - \frac{1}{2} m v^{2} = \frac{1}{2} m \cdot \frac{ħ^{2}}{m^{2} r^{2}} = \frac{ħ^{2}}{2 m {[r_{B} + δ]}^{2}} = \frac{ħ^{2}}{2 m {r_{B}}^{2}} {(1 + \frac{δ}{r_{B}})}^{- 2} = \frac{ħ^{2}}{2 m {r_{B}}^{2}} (1 - \frac{2 δ}{r_{B}})

Total energy = - \frac{e^{2}}{4 {πε}_{0} r_{B}} + \frac{ħ^{2}}{2 m r_{B}^{2}} [1 + 2 {λr}_{B}]

= - 27.2 + 13.6 [1 + 2 {λr}_{B}] eV

Change in energy = 1.36 \times 2 {λr}_{B} eV = 27.2 {λr}_{B} eV

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