Q: In a cathode ray tube, a p.d of 3000 v is maintained between the deflector plates whose separation is 2 cm. A magnetic field of 2.5 × 10^{-3} T at right angles to the electric field gives no deflection of electron beam, which received an initial acceleration by a p.d of 10,000 V. the e/m of electron is

Sol: V_{1} is p.d between the deflector plates and V_{2} be the potential difference through which electrons are

accelerated .

$\large \frac{1}{2} m v^2 = q V_2 $

$\large \frac{1}{2} m (\frac{V_1}{d B})^2 = q V_2 $

$\large \frac{q}{m} = \frac{V_1^2}{2 d^2 B^2 V_2} $

$\large \frac{q}{m} = \frac{9 \times 10^6}{2 \times 10^{-4}\times 6.25 \times 10^{-6}\times 10^4} $

= 1.8 × 10^{11} C/kg