MCQ
In a Thomson set-up for the determination of $\mathrm{e} / \mathrm{m}$, electrons accelerated by $2.5 \mathrm{kV}$ enter the region of crossed electric and magnetic fields of strengths $3.6 \times 10^4 \mathrm{Vm}^{-1}$ and $1.2 \times 10^{-3} \mathrm{~T}$ respectively and go through undeflected. The measured value of $e / m$ of the electron is equal to
  • A
    $1.0 \times 10^{11} \mathrm{C}-\mathrm{kg}^{-1}$
  • B
    $1.76 \times 10^{11} \mathrm{C}-\mathrm{kg}^{-1}$
  • $1.80 \times 10^{11} \mathrm{C}-\mathrm{kg}^{-1}$
  • D
    $1.85 \times 10^{11} \mathrm{C}-\mathrm{kg}^{-1}$

Answer

Correct option: C.
$1.80 \times 10^{11} \mathrm{C}-\mathrm{kg}^{-1}$
$ \frac{e}{m}=\frac{E^2}{2 V B^2}=\frac{\left(3.6 \times 10^4\right)^2}{2 \times 2.5 \times 10^3 \times\left(1.2 \times 10^{-3}\right)^2} $
$ =1.8 \times 10^{11} \mathrm{C} / \mathrm{kg}$

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