In a photoelectric cell, for the incident radiation of wavelength , the fastest electron has speed 'v'. If the wavelength of incident radiation is changed to 3 , the speed of the fastest emitted electron is

Step-by-step solution

Using Einstein's photoelectric equation: 1 2 mv^2 = hc - When the wavelength is changed to 3 , the new maximum kinetic energy is: 1 2 mv'^2 = 3hc - Substituting hc = 1 2 mv^2 + into the second equation: 1 2 mv'^2 = 3 ( 1 2 mv^2 + ) - 1 2 mv'^2 = 3 ( 1 2 mv^2 ) + 2 Since the work function > 0 , we get: 1 2 mv'^2 > 3 ( 1 2 mv^2 ) v'^2 > 3v^2 v' > v 3 Thus, the speed of the fastest emitted electron is greater than v( 3 ) . Answer: greater than v( 3 )