Question

A particle having mass 1 g and electric charge 10⁻⁸ C travels from a point A having electric potential 600 V to the point B having zero potential. What would be the change in its kinetic energy ?(A) -6 x 10⁻⁶ erg (B) -6 x 10⁻⁶ J(C) 6 x 10⁻⁶ J (D) 6 x 10⁻⁶ erg

Answer 1

Lambda = 5600 A,

Energy of one photon is E = h\nu = h*c/\lambda = 6.63* 10^-34 * 3* 10^8/(5600* 10^-10) = 3.616 * 10^-19 J, A 100 watt bulb requires 100 joule of energy per second.

Energy radiated by the bulb as visible light per second = 5% of 100, = 5 Joule.

Number of photons emitted per second= Energy radiated per sec/ energy of one photon = 5/(3.616 * 10^ -19) = 1.38 * 10^19