Question

What is the kinetic energy of the electron in the \( n \)th level, moving in a plane under the influence of a magnetic field \( B \)? \([ m\text{ - mass of electron},\; h\text{ - Planck's constant},\; e\text{ - electronic charge}]\)

• A. \( \frac{heB}{4\pi m n} \)
• B. \( \frac{nheB}{4\pi m} \)
• C. \( \frac{nheB}{2\pi m} \)
• D. \( \frac{heB}{2\pi m n} \)

Hint:

Use \( evB = \frac{mv^2}{r} \) and Bohr quantization together to derive energy in magnetic field problems.

Answer 1

The Correct Option is B

Step 1: Magnetic force provides centripetal force.
\[ evB = \frac{mv^2}{r} \]

Step 2: Simplify the equation.
\[ v = \frac{eBr}{m} \]

Step 3: Apply quantization condition.
\[ mvr = \frac{nh}{2\pi} \]

Step 4: Substitute value of \( v \).
\[ m \left(\frac{eBr}{m}\right) r = \frac{nh}{2\pi} \]
\[ eBr^2 = \frac{nh}{2\pi} \]

Step 5: Find \( r^2 \).
\[ r^2 = \frac{nh}{2\pi eB} \]

Step 6: Write kinetic energy.
\[ K = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{eBr}{m}\right)^2 \]
\[ K = \frac{e^2 B^2 r^2}{2m} \]
Substitute \( r^2 \):
\[ K = \frac{e^2 B^2}{2m} \cdot \frac{nh}{2\pi eB} \]

Step 7: Simplify.
\[ K = \frac{nheB}{4\pi m} \]
\[ \boxed{\frac{nheB}{4\pi m}} \]