Question

The orbiting speed $v_n$ of $e^{-}$ in the $nth$ orbit in the case of positronium is $x-fold$ compared to that in $nth$ orbit in a hydrogen atom, where $x$ has the value

• A. $1$
• B. $\sqrt{2}$
• C. $1/\sqrt{2}$
• D. $2$

Answer 1

The correct option is D $2$

*positronium is system consisting of electron and its antiparticle i.e electron bound together into an exotic atom it has charge magnitude equal to the electron and

it is a reduced mass, here Me is mass of electron so the mass of positronium is 1/2 of electron

The orbiting speed $V_n$ is given by:-

$V_n=Z{e}^2/4\pi ϵnh$

and $\frac{e^2}{4\pi {ϵ}_0{r}^2}=\frac{m{v}^2}{r}$

& $mvr=\frac{nh}{2\pi }$

So, $n\propto m$

so by applying above formula

$V_e=xVp$

so we get value of x=2.