Question

If the mass of an electron is doubled, then find the new radius (in $\mathring{A})$ of $1^{st}$ orbit of $H-$atom?

Answer 1

By Bohr hypothesis, the radius of the $n^{th}$ orbit is given as $R_n=\frac{n^2{h}^2}{4{\pi }^{2}m{Ze}^2}$

In case of Hydrogen, first orbit, n=1 and Ze=1 Hence, $R=\frac{h^2}{4{\pi }^{2}m}$

If the mass is doubled, then $R=\frac{h^2}{4{\pi }^2\left(2m\right)}$ which clearly indicates that the radius is halved.