Find the frequency of revolution of an electron with an energy of $100 e V$ in a uniform magnetic field of magnitude $35.0 μ T$ .

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Solution

The frequency of revolution is
$f = \frac{B q}{2 π m_{e}}$
$= \frac{(35.0 \times 10^{- 6} T) (1.60 \times 10^{- 19} C)}{2 π (9.11 \times 10^{- 31} k g)}$
$= 9.78 \times 10^{5} H z$ .

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Find the frequency of revolution of an electron with an energy of 100eV in a uniform magnetic field of magnitude 35.0μT.

The frequency of revolution isf=Bq2πme=(35.0×10−6T)(1.60×10−19C)2π(9.11×10−31kg)=9.78×105Hz.

Find the frequency of revolution of an electron with an energy of $100 e V$ in a uniform magnetic field of magnitude $35.0 μ T$ .

The frequency of revolution is
$f = \frac{B q}{2 π m_{e}}$
$= \frac{(35.0 \times 10^{- 6} T) (1.60 \times 10^{- 19} C)}{2 π (9.11 \times 10^{- 31} k g)}$
$= 9.78 \times 10^{5} H z$ .