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Q 12.12  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.

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As per the bohrs model

 m_{e}v_{n}r_{n}=\frac{nh}{2\pi}                        (i)

If the proton and the electron were bound only by the gravity the gravitational force between them will provide the centripetal force required for circular motion

\frac{m_{e}v_{n}^{2}}{r_{n}}=\frac{Gm_{e}m_{p}}{r_{n}^{2}}                   (ii)

From equation (i) and (ii) we can calculate that the radius of the ground state (for n=1) will be

\\r_{1}=\frac{h^{2}}{4\pi Gm_{p}m_{e}^{2}}\\ r_{1}=\frac{(6.62\times 10^{-34})^{2}}{4\pi \times 6.67\times 10^{-11}\times 1.67\times 10^{-27}\times (9.1\times 10^{-31})^{2}}

r_{1}\approx 1.2\times 10^{29}\ m

The above value is larger in order than the diameter of the observable universe. This shows how very weak the gravitational forces of attraction are as compared to electrostatic forces.

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