A photosensitive metal is exposed by electromagnetic radiation of frequency greater than threshold frequency. It is observed that after 5 seconds mass of metal decrease by 0.091 pg. How many photoelectrons are emitted from the surface?

Solution

The mass of electron is 9.1 x 10^–31kg.

Decrease in the mass of metal = 0.091 pg =

0.091 \times 10^{- 12} g

Thus, emitting one electron shall decrease the mass of metal by a mass equal to mass of electron.

Therefore, the number of electrons emitted is,

n = \frac{Δm}{m} = \frac{0.091 x 10^{- 12}}{9.1 x 10^{- 28}} = 10^{14}

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