If maximum velocity with which an electron can be emitted from a photo cell is `4 xx 10^(8) cm//sec` , the stopping potential is (mass of electron ` = 9 xx -31 kg`)

If maximum velocity with which an electron can be emitted is 4xx10^8 cm s^-1 , then find the stopping potential V_o (mass of electron =9xx10^(-31) kg)

Electrons are emitted from photo-cell with maximum speed of 4xx10^(8) cm/s .Its stopping potential will be….. (Mass of electron =9xx10^(-31) kg.

If maximum K.E. of electron emitted by a photo cell is 4 eV, what is the stopping potential?

Maximum velocity of photoelectron emitted is 4.8 m//s . If e//m ratio of electron is 1.76 xx 10^11 (kg^-1) then stopping potential is given by

The maximum velocity of ejected photoelectrons 2 xx 10^6 m/sec and (e/m) of electrons is 1.8 xx 10^11 coul/kg. The stopping potential in volts is

The maximum velocity of photoelectronos emitted hy a photoemitter is 1.8 xx 10^(6) m//sec . Taking e//m = 1.8 xx 10^(11) C/kg for electrons, the stopping potential of the emitter is:

Maximum velocity of the photoelectron emitted by a metal is 1.8 xx 10^(6) ms^(-1) . Take the value of specific charge of the electron is 1.8 xx 10^(11) C kg^(-1) . Then the stopping potential in volt is

The velocity of electron in hydrogen atom is 7.29 xx 10^(7) cm/sec The potential energy of that electron is