If kinetic energy of electron is double then its radius becomes?

Question

3210 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-11T06:37:12

$\frac{1}{\sqrt{2}}$ $1/sqrt(2)$ times

For,an electron moving round the nucleus in its orbit,its angular momentum is conserved.

That is, $m v r = C$ $mvr=C$ (where, $m$ $m$ is the mass, $v$ $v$ is the velocity and $r$ $r$ is the radius , $C$ $C$ is a constant)

Now,mass of electron being constant we can say, $r \propto \frac{1}{v}$ $r prop 1/v$

Now,kinetic energy of an electron can be written as $K = \frac{1}{2} m v^{2}$ $K=1/2 mv^2$

So, $K \propto v^{2}$ $K prop v^2$

so,if kinetic energy is doubled,its velocity will become $\sqrt{2}$ $sqrt(2)$ times than the initial.

So,its radius being inversely proportional to its velocity in an orbit,it will be $\frac{1}{\sqrt{2}}$ $1/sqrt(2)$ times the initial value!