Questions & Answers

Question

Answers

(A) $ n\left( \dfrac{hqB}{4\pi m} \right) $

(B) $ n\left( \dfrac{hqB}{8\pi m} \right) $

(C) $ n\left( \dfrac{hqB}{\pi m} \right) $

(D) $ n\left( \dfrac{hqB}{2\pi m} \right) $

Answer

Verified

92.1k+ views

The magnetic Lorentz force provides the necessary centripetal force. From here, value of $ \dfrac{v}{r} $ can be calculated

By multiplying these two values found, we can find the value of $ {{v}^{2}} $ and hence kinetic energy using the formula $ \dfrac{1}{2}m{{v}^{2}} $.

Bohr’s quantization condition states that angular momentum of an electron is an integral multiple of $ \dfrac{h}{2\pi } $

$ \text{L=}\dfrac{n\text{ }h}{2\pi } $

As $ \alpha =mvr $ , so

$ mvr=\dfrac{nh}{2\pi } $

$ vr=\dfrac{nh}{2\pi m} $ …… …… …… (1)

Now, the magnetic Lorentz force provides the necessary centripetal force.

So, $ \dfrac{m{{v}^{2}}}{r}=qvB $

On rearranging, we get

$ \dfrac{v}{r}=\dfrac{qB}{m} $ …… …… …… (2)

On multiplying equations (1) and (2) we get

$ vr\times \dfrac{v}{r}=\left( \dfrac{nh}{2\pi m} \right)\times \left( \dfrac{qB}{m} \right) $

$ {{v}^{2}}=\dfrac{n\text{ }h\ q\text{ }B}{2\pi \text{ }{{m}^{2}}} $

Now energy of the charged particle is given by:

K.E $ =\dfrac{1}{2}m{{v}^{2}} $

$ =\dfrac{1}{2}m\left( \dfrac{n\text{ }h\text{ }q\text{ }B}{2\pi {{m}^{2}}} \right) $

$ \text{E=n}\left( \dfrac{hqB}{4\pi m} \right) $.

Lorentz force is the force exerted on a charged particle q moving with velocity v through an electric field E and moving magnetic field B.

The entire electromagnetic force F on the charged particle is called the Lorentz force and is given by

$ \text{F=}qE+qv\times B $

The first term is contributed by the electric field. The second term is the magnetic field. The second term is the magnetic field and has a direction perpendicular to both the velocity and the magnetic field.