RRB Technician Grade I 20 Dec 2024 Shift 1 Paper

Concept:An electron accelerated from rest through a potential difference $V$ gains kinetic energy equal to the work done by the electric field.Explanation:Since the electrons leave the cathode with zero velocity, their initial kinetic energy is zero.The work done on an electron by the accelerating voltage is $W = eV$.By energy conservation, this work equals the kinetic energy gained:$eV = \frac{1}{2}mv^2$.Rearranging for velocity:$v = \sqrt{\frac{2eV}{m}}$.Given: $V = 182$ V, $e = 1.6 \times 10^{-19}$ C, $m = 9.1 \times 10^{-31}$ kg.Substitute the values:$v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 182}{9.1 \times 10^{-31}}}$.Simplify the numerator:$2 \times 1.6 \times 10^{-19} \times 182 = 5.824 \times 10^{-17}$.So, $v = \sqrt{\frac{5.824 \times 10^{-17}}{9.1 \times 10^{-31}}}$.$v = \sqrt{6.4 \times 10^{13}}$.This gives $v = \sqrt{64 \times 10^{12}} = 8 \times 10^6$ m/s.Answer:The maximum velocity of the electron beam is $8 \times 10^6$ m/s, which matches option B.