Concept:In J.J. Thomson's experiment (or in a standard mass spectrometer setup), an electron is accelerated through a potential difference \( V \) and then enters a perpendicular magnetic field \( B \), moving in a circle of radius \( r \).
Formula:$$ \frac{1}{2}mv^2 = eV \quad \text{and} \quad evB = \frac{mv^2}{r} $$
Solution:- From the magnetic force providing centripetal force: \( v = \frac{eBr}{m} \).
- Substitute this velocity \( v \) into the kinetic energy equation: \( \frac{1}{2}m \left( \frac{eBr}{m} \right)^2 = eV \).
- Simplify the square: \( \frac{1}{2}m \left( \frac{e^2 B^2 r^2}{m^2} \right) = eV \).
- Cancel \( m \) and rearrange to isolate \( \frac{e}{m} \): \( \frac{e B^2 r^2}{2m} = V \implies \frac{e}{m} = \frac{2V}{B^2 r^2} \).
Why other options are incorrect:Option B uses the square of \( V \) incorrectly. Options C and D mix up the dimensions and fail to account for the kinetic energy relationship.
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