Question

A photocell, such as that illustrated in Figure 6.7(b), is a device used to measure the intensity of light. In a certain experiment, when light of wavelength 590 nm is directed on to the photocell, electrons are emitted at the rate of 5.6 ✕ 10^-13 C/s. Assume that each photon that impinges on the photocell emits one electron.

How many photons per second are striking the photocell?

How much energy per second is the photocell absorbing?

Answer 1

The charge of an electron = 1.6*10^-19 Coulomb (C)

i.e. 5.6*10^-13 C = (5.6*10^-13/1.6*10^-19) electrons = 3.5*10⁶ electrons

Here, each photon that strikes on the photo cell emits one electron.

Hence, the no. of photons striking the photocell per second (N) = 3.5*10⁶

The energy of N-photons can be calculated as follows.

E = Nhc/, where h is the Planck's constant = 6.626*10^-34 Js, c is the velocity of light = 3*10⁸ m/s and = 590 nm

i.e. E = (3.5*10⁶ * 6.626*10^-34 Js * 3*10⁸ m/s)/(590*10^-9 m)

i.e. E = 1.18*10^-12 J