1.

The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photo electrons emitted?

Answer»

V0 = 1.5 V
∴ Maximum kinetic energy of the emitted electrons,

\(\frac{1}{2}mV^2_{max} = eV_0= 1.5j= 15eV\)



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