Question

23. If elm of electron is 1.76 * 1011 C kg and the stopping
potential is 0.71 V, then the maximum velocity of the
photoelectron is
(a) 150 km -
(b) 200 km s-1
(c) 500 km s-
(d) 250 km s-​

Answer 1

Explanation:

eV=1/2 mv^2

(e/m)V=v^2/2

v^2=2(e/m)V

v^2=2×1.76×10^11×0.71

v=4.99×10^5 m/s=5×10^5 m/s=500 ×10^3 m/s=500 km/s