when radiation is incident on photoelectron emitter the stopping potential is found to be 9 V if e/m for the electron is 1.8x1011 C/kg the maximum velocity of the ejected electron is
Hi Siddhant, recall 1/2 * m * v2max = e V
Given e/m = 1.8 * 10^11 C/kg and V = 9 V
So rearranging we get v2max = 2 * e/m * V = 2 * 1.8 * 10^11 * 9
v2max = 18^2 * 10^10
So vmax = 18 * 10^5 m/s = 1.8 * 10^6 m/s
Given e/m = 1.8 * 10^11 C/kg and V = 9 V
So rearranging we get v2max = 2 * e/m * V = 2 * 1.8 * 10^11 * 9
v2max = 18^2 * 10^10
So vmax = 18 * 10^5 m/s = 1.8 * 10^6 m/s