Question

In a photoelectric effect experiment, irradiation of metal with the light of frequency $5.2 \times {10^{14}}{s^{ - 1}}$ yields electrons with maximum kinetic energy $1.3 \times {10^{ - 19}}$. Calculate the threshold frequency $\left( {\nu _0} \right)$ for the metal.

Answer 1

Hint: We know that the ejection of electrons from a metal surface when a beam of light of certain minimum frequency is allowed to strike the metal surface in a vacuum is known as the photoelectric effect.

Formula Used:
$K.E = h\left( {\nu - {\nu _0}} \right)$
Where K.E is the kinetic energy
${\nu _0}$ is the threshold frequency
$\nu $ is the frequency

Complete step by step solution:
We know that from the properties of the photoelectric effect that if the frequency of light is more than the threshold frequency then some of the energy gets consumed to eject the electron from an atom and the extra energy (excess to that of threshold energy) renders the electron with kinetic energy $\dfrac{1}{2}m{v^2}$
i.e., $Total\;energy\;\left( E \right) = I.E + K.E$
where I.E is the ionisation energy required to knock the electron and K.E is the kinetic energy of the electron.
It can be written as
$h\nu = h{\nu _0} + \dfrac{1}{2}m{v^2}$
Where,$h$ is the planck's constant
${\nu _0}$ is the threshold frequency
$\nu $ is the frequency
We have given that
$Frequency\;of\;metal = 5.2 \times {10^{14}}{s^{ - 1}}$
$Kinetic\;energy = 1.3 \times {10^{ - 19}}$
Rewriting the equation,
$\Rightarrow {{h}}{{{\nu }}_0} = {{h\nu }} - {{K}}.{{E}}$
$\Rightarrow {{{\nu }}_0} = {{\nu }} - \dfrac{{{{K}}.{{E}}}}{{{h}}}$
Also, we know that planck's constant is $6.626 \times {10^{ - 34}}$
Substituting the values we get,
$\Rightarrow {\nu _0} = 5.2 \times {10^{14}} - \dfrac{{1.3 \times {{10}^{ - 19}}}}{{6.626 \times {{10}^{ - 34}}}}$
On solving, we get
$\Rightarrow {\nu _0} = 3.24 \times {10^4}{s^{ - 1}}$

That is, the threshold frequency for the metal is $3.24 \times {10^4}{s^{ - 1}}$

Note:
When the number of photoelectrons ejected is proportional to the intensity of incident radiation i.e., an increase in the intensity of incident radiation increases the rate of emission, not the energy of the photoelectron.
Also, we can see that the alkali metals having lower ionization energy readily show the photoelectric effect. Caesium having lowest ionisation energy, shows this effect in the best way. Hence it is used in photoelectric cells.