Question

If an electron in $H$-atoms jumps from one orbit to other its angular momentum doubles. If the new radius is ${}^{\prime }{x}^{\prime }$ times the old radius, then find $x$ ?

Answer 1

Angular momentum $=\frac{nh}{2\pi }\left\{fornthshall\right\}$

$\therefore$ Let shall be $n$ and $n^1$

comparing angular momentum

$2\times \frac{nh}{2\pi }=\frac{n^{\prime }}{2\pi }$

$2=\frac{n^{\prime }}{n}$

$radius=r=\frac{0.53n^2}{z}=0.53n^2$

$r^{\prime }=0.53(n^{\prime }{)}^2.$

Now $x=\frac{r^{\prime }}{r}=\frac{0.53(n^{\prime }{)}^2}{0.53n^2}=(\frac{n^{\prime }}{n}{)}^2$

$x=4$