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 560 nm is directed on to the photocell, electrons are emitted at the rate of 9.6 ✕ 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?
photons/s
How much energy per second is the photocell absorbing?J/s

Answer 1

A)

We know that

1 coulomb = 6.25 x 10¹⁸ electrons

So, 9.6 x 10^-13 coulomb = (9.6 x 10^-13) (6.25 x 10¹⁸) electrons
​    = 6000000 electrons
                             = 6.00 x 10⁶​ electrons

Hence, 6.00 x 10⁶​ number of photons/second.

B)

Each photon has energy

E = h c /

where
h = plank's constant = 6.63 x 10^-34 J s
c = speed of light = 3.0 x 10⁸ m/s and
= wavelength = 560 nm = 5.60 x 10^-7 m,

Substituing the above values, we get;

E = [ (6.63 x 10^-34 J s ) (3.0 x 10⁸ m/s) ] / (5.60 x 10^-7 m)
   = 3.55 x 10^-19 J

So, the energy per second is the photocell absorbing​ = 3.55 x 10^-19 J/s