Question

A monochromatic light source of intensity 5 mW emits $8\times {10}^{15}$ photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of the metal.

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Answer 1

Power of light source = 5mW = $5\times {10}^{-3}$ W

E = Energy delivered by light source per second = $5\times {10}^{-3}$ J …(i)

n = No. of photons coming out = $8\times {10}^{15}$ per sec. …(ii)

What is the energy of each photon.

We can find that right ?

We know

E = nhv where hv is energy of each photon from (i) and (ii)

$5\times {10}^{-3}=8\times {10}^{15}\times$ hv

$hv=\frac{5\times {10}^{-3}}{8\times {10}^{15}}$joules …(iii)

Let’s assume the work function is $\Phi$ so after photon hits the electron, it transfers hv of energy. Work function is φ so that much energy is spent in coming out. Assume the electron is the most loosely bound and it doesn’t spend any energy in collisions. So when it comes out, it has maximum energy. Stopping potential is 2 V so the energy spent to stop this electron is $1.6\times {10}^{-19}\times 2$J.

Energy is conserved

When electron came out it had kinetic energy of hv-$\Phi$, and 0 potential energy. When it stops it has 0 kinetic energy but $1.6\times {10}^{-19}\times 2$ J of potential energy.

$\Rightarrow hv–\Phi =1.6\times {10}^{-19}\times 2$

$\Phi =hv–1.6\times {10}^{-19}\times 2$

from (iii)

$\Phi =\frac{5\times {10}^{-3}}{8\times {10}^{15}}1.6\times {10}^{-19}\times 2$

$=6.25\times {10}^{-19}–3.2\times {10}^{-19}$

$\Phi =3.05\times {10}^{-19}$ J

in ev $\Phi$ $=\frac{3.05\times {10}^{-19}}{1.6\times {10}^{-19}}$
= 1.9 eV