Name: Bohr Model of the Hydrogen Atom
Uploaded: 2019-07-19
Duration: 4 min 50 s
Description: In this video, we will discuss the Bohr model of the hydrogen atom and electron transition in the hydrogen atom. Get in this to explore more.

Question

The inverse square law in electrostatics 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}

, force would be modified to

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

where

λ = m_{p} c / h

and

h = \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.

Accepted Answer

Mass of photon $= 9.1 \times 10^{- 31} \times 10^{- 6} kg$ $= 9.1 \times 10^{- 37} kg$

Wavelength associated with a photon $= \frac{h}{m_{p} c}$

$λ = \frac{6.62 \times 10^{- 34}}{9.1 \times 10^{- 37} \times 3 \times 10^{8}}$ $= \frac{6.62}{9.1 \times 3} \times 10^{- 34 + 37 - 8} = 2.4 \times 10^{- 7} m > > r_{A}$

Bohr radius , $r_{A} = 0.53 \times 10^{- 10} m$

$λ r_{A} < < 1$

Given ,

$| F | = \frac{e^{2}}{(4 {πε}_{0}) r^{2}} [\frac{1}{r^{2}} + \frac{λ}{r}] . e x p (- λ r) - - - - (1) | F | = \frac{d U}{d r}$

On solving we get,

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

$m v r = \frac{h}{2 π} = h$ or $v = \frac{h}{m r} . . .$

$[∵ F = \frac{r^{2}}{(4 {πε}_{0}) r^{2}} [\frac{1}{r^{2}} + \frac{λ}{r}] e^{- λ r} given,] T a k i n g e^{- λ r} \approx 1 \frac{m v^{2}}{r} = \frac{e^{2}}{4 {πε}_{0}} [\frac{1}{r^{2}} + \frac{λ}{r}]$

$\frac{m}{r} \times \frac{h^{2}}{m^{2} r^{2}} = \frac{e^{2}}{4 {πε}_{0}} [\frac{1 + λ r}{r^{2}}]$

$\frac{h^{2}}{m r} = \frac{e^{2}}{4 {πε}_{0}} (1 + λ r)$

$\frac{h^{2} 4 {πε}_{0}}{m e^{2}} = (r + λ r^{2})$

$\frac{h^{2}}{m} = \frac{e^{2}}{4 {πε}_{0}} r_{A} (r_{A} = r + λ r^{2})$

Let $r = r_{A} + δ$

$r_{A} = (r_{A} + δ) + λ {(r_{A} + δ)}^{2} r_{A} = (r_{A} + δ) + λ {({r^{2}}_{A} + δ^{2} + 2 r_{A} δ)}^{2}$

$0 = δ + λ {r_{A}}^{2} + 2 r_{A} δ λ$ (neglecting small term $δ^{2}$ )

$0 = δ + 2 r_{A} δ λ + λ {r_{A}}^{2}$

$\Rightarrow δ [1 + 2 r_{A} λ] = - λ {r_{A}}^{2}$

$δ = \frac{- λ {r_{A}}^{2}}{(1 + 2 r_{A} λ)} = - λ {r_{A}}^{2} (1 + 2 r_{A} λ)^{- 1}$

$δ = - λ {r_{A}}^{2} {[1 + 2 r_{A} λ]}^{- 1} = - λ {r^{2}}_{A} + 2 {r^{3}}_{A} λ^{2}$

$∴ λ$ and $r_{A} < < 1$ so ${r_{A}}^{3} λ^{2}$ is very small so by neglecting it we get

$[δ = - λ {r_{A}}^{2}]$

$U (r) = \frac{- e^{2} e^{- λ (r_{A} + δ)}}{4 π \in_{0} (r_{A} + δ)}$
$= \frac{- e^{2}}{4 π \in_{0} r_{A}} e^{- λ r} {(1 + \frac{δ}{r_{A}})}^{- 1} ∵ (δ^{2})$

$= \frac{- e^{2} e^{- 2 r}}{4 π \in_{0} r_{A}} (1 + \frac{δ}{r_{A}})$

$V (r) = - 27.2 eV$ remains unchanged

$K . E . = \frac{1}{2} m v^{2} = \frac{1}{2} m {(\frac{h}{m r})}^{2} = \frac{1}{2} \frac{h^{2}}{m r^{2}} (From, v = \frac{h}{m r})$

$= \frac{- h^{2}}{2 m (r_{A} + δ)^{2}} = \frac{- h^{2}}{2 m {r_{A}}^{2} {(1 + \frac{δ}{r_{A}})}^{2}} = \frac{h^{2}}{2 m {r_{A}}^{2}} (1 - \frac{2 δ}{r_{A}})$ $= \frac{h^{2}}{2 m r_{A}} (1 - \frac{- λ {r_{A}}^{2}}{r_{A}}) = \frac{h^{2}}{2 m r_{A}} (1 + 2 λ r_{A})$

$= 13.6 e V (1 + 2 λ r_{A})$

Total energy $- \frac{- r^{2}}{4 π \in_{0} r_{A}} + \frac{h^{2}}{2 m {r_{A}}^{2}} (1 + 2 λ r_{A})$

$- [- 27.2 + 13.6 (1 + 2 λ r_{A})] e V$

$= - 27.2 + 13.6 + 27.2 λ r_{A} e V$

Total $E = - 13.6 + 27.2 λ r_{A}$

Change in energy $= 13.6 + 27.2 λ r_{A} - (13.6) = 27.2 λ r_{A} e V$

Important Questions on Atoms

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