In a photocell, ultraviolet light provides enough energy to some electrons in ba

Question

In a photocell, ultraviolet light provides enough energy to some electrons in barium metal to eject them from the surface at high speed. See Fig. 23-36. To measure the maximum energy of the electrons, another plate above the barium surface is kept at a negative enough potential that the emitted electrons are slowed down and stopped, and return to the barium surface. If the plate voltage is -3.02V when the fastest electrons are stopped, what was the speed of these electrons when they were emitted?

Explanation / Answer

So energy of emitted electron = eV = 3.02 eV =3.02*1.6*10-19 J = 4.832 *10-19 J (As at that time , electron is at rest , So kinetic energy iszero )
So this energy should be equal to maximum kinetic energy, So mv2/2 = 4.832 *10-19 J m= mass of electron = 9.1*10-31 kg v=1.03*106 m/s

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