Question

Express the velocity of an electron in the n-th orbit of the hydrogen atom, in terms of the principal quantum number n itself.

• A. $v=\left(\frac{e^2}{2h{ϵ}_0}\right)\frac{1}{n^2}$
• B. $v=\left(\frac{e}{2{ϵ}_0}\right)\frac{1}{n}$
• C. $v=\left(\frac{e^2}{2h{ϵ}_0}\right)\frac{1}{n}$
• D. $v=\left(\frac{e^2}{2h{ϵ}_0}\right)n^2$

Answer 1

The correct option is C $v=\left(\frac{e^2}{2h{ϵ}_0}\right)\frac{1}{n}$

Seeing how the Rutherford model predicted an unstable atom, Bohr postulated, as we know, that for certain discrete values of the angular momentum $(L=\frac{nh}{2\pi }),$ the electron goes around in stable orbits without radiating. For a point mass m going around in a circular orbit of radius r with speed v, we know that the angular momentum is L=mvr. Combining –
$mvr=\frac{nh}{2\pi }$
$\Rightarrow v=\frac{nh}{2\pi mr}.$ …(1)

Recall that for Z=1 (hydrogen), we had expressed the radius r in terms of the quantum number n as –
$r=\left(\frac{h^2{ϵ}_0}{\pi m{e}^2}\right)n^2$ …(2)

Substituting r from (2) in (1) we obtain –
$v=\left(\frac{e^2}{2\pi {ϵ}_0}\right)\frac{1}{n}.$

How would you interpret the constant in the brackets? Convince yourself that it is basically the speed of the electron in the ground state (n=1) of the hydrogen atom.