The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10^–40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

Solution

The radius of the first orbit of hydrogen atom in Bohr's model is given by,

r = \frac{ε_{0} h^{2}}{π {me}^{2}} Multiplying and dividing by 4 π r = \frac{4 {πε}_{0}}{e^{2}} (\frac{h^{2}}{4 π^{2} m})

Therefore,

Radius,

r = \frac{n^{2} h^{2}}{4 π^{2} {mkZe}^{2}}

[here k = \frac{1}{4 {πε}_{0}} Z = 1, n = 1]

If electrostatic force

\frac{1}{4 {πε}_{0}} . \frac{e^{2}}{r^{2}}

is replaced by gravitational force

\frac{GMm}{r^{2}},

we put GMm in place of

\frac{e^{2}}{4 {πε}_{0}}

in above expression.

Hence, radius of first orbit under gravitational force,

r_{G} = \frac{1}{GMm} . \frac{h^{2}}{4 π^{2} m}

= \frac{h^{2}}{4 π^{^{2}} {GMm}^{2}}

, where

[M = mass of proton m = mass of electron]

\Rightarrow

r_{G} = \frac{(6.26 \times 10^{- 34})^{2}}{4 \times (3.14)^{2} (6.67 \times 10^{- 11}) \times (1.672 \times 10^{- 27}) \times (9.1 \times 10^{- 31})^{2}}

= \frac{6626 \times 6626 \times 10^{- 74}}{4 \times 3.14 \times 3.14 \times 667 \times 16724 \times 19 \times 19 \times 10^{- 112}} = 1.21 \times 10^{29} m .

The radius of the first orbit, when electron and proton are bounded together under the influence of graviatational force, turns out to be larger than the size of the universe.

Some More Questions From Wave Optics Chapter

Answer the following questions, which help you understand the difference between Thomson's model and Rutherford's model better.
Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson's model much less, about the same, or much greater than that predicted by Rutherford's model?

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