For hydrogen like atom, find velocity from potential energy

PE of system is given by $$\mathrm{U(r) = \frac{-Ke²}{r^3}}$$ assume bohr model to be valid Find velocity

For this I used $\mathrm{\frac{-PE}{2} =KE}$ and then used $$\frac{mv^2}{2} =\frac{Ke^2}{2r^3}$$ and $$mvr=\frac{nh}{2\pi}$$

However, I am getting the answer as $$v=\frac{n^3h^3}{16Ke^2\pi^3m^2}$$ but the answer given has a factor of 24 rather than 16. I don't know where I'm going wrong.

• Your question could really benefit from some formatting. As for the -PE/2 =KE part: where did you take that from? It looks suspiciously like the special case of virial theorem for Coulomb-like forces, which your force is not. – Ivan Neretin Mar 14 '18 at 7:35
• my concern is -PE/2 = KE i think it applies to usual atoms as there PE is proportional to 1/r. here it might change – Karmanya GB Mar 14 '18 at 11:21

The fault in your solution is the part where you assume $\mathrm{\frac{-PE}{2} =KE}$ this is true only if the force follows inverse square law.

In this particular case we need to find the force first by using

$$F =\frac{-dU}{dr}$$

Which gives us

$$F =\frac{-3Ke^2}{r^4}$$

Now this force is central and provides centripetal force

$$\frac{mv^2}{r} =\frac{3Ke^2}{r^4}$$

This along with $$mvr=\frac{nh}{2\pi}$$

gives us $$v=\frac{n^3h^3}{24Ke^2\pi^3m^2}$$

The solution of the questions is given in steps in the figure.