Question

The angular momentum of an electron in a given stationary state can be expressed by the equation MVR=nh/2π ​

Answer 1

Answer:

नही .............आता

Answer 2

Given:

In the question given that $\bold{MVR = \frac{nh}{2\pi} }$

To Find:

Proof of the given expression $\bold{MVR = \frac{nh}{2\pi} }$.

Solution:

Before going further we have to know about the given equation. It is Bohr's postulate of quantization of angular momentum.

To proof this equation we will take help of de-Broglie wavelength concept.

Now let, an electron moving in $n^{th}$ circular orbit of radius R then

                                                      $\bold{2\pi R = n\lambda}$        …..(1)

We know that, de-Broglie wavelength, $\bold{\lambda = \frac{h}{p} }$, where h = Plank's constant, p= momentum of electron.

And   $\bold{\lambda = \frac{h}{MV} }$, (∵ momentum p = MV, M = mass of electron, V = speed of electron)

Using value of $\lambda$,

(1) =>                                              $\bold{2\pi R = \frac{nh}{MV} }$

                                              $=>\bold{ MVR = \frac{nh}{2\pi} }$

Hence, in a given stationary state angular momentum of an electron is$\bold{ MVR = \frac{nh}{2\pi} }$