Question

Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 $\mathop A\limits^ \circ$.
Calculate threshold frequency (${{v}_{o}}$ ) and work function of the metal.

Answer 1

There is a formula to calculate the work function and threshold frequency of a metal and they are as follows.
${{v}_{o}}=v=\dfrac{c}{\lambda }$
Here ${{v}_{o}}$ = threshold frequency
c = velocity of light
$\lambda$ = wavelength of the light
Kinetic energy = $h{{v}_{o}}$
Here h = planck's constant
${{v}_{o}}$ = threshold frequency

Complete answer:
- In the question it is asked to calculate the work function and threshold frequency of the metal by using the given data.
- In the question it is given that the wavelength of the radiation is 6000 $\mathop A\limits^ \circ$.
- We know that if kinetic energy is zero then the energy of the incident radiation is equal to the work function of the metal.
- Therefore we can calculate the work function of the metal easily.
- First we have to calculate the threshold frequency of the radiation by using the given wavelength and it is as follows.
${{v}_{o}}=v=\dfrac{c}{\lambda }$
Here ${{v}_{o}}$ = threshold frequency
c = velocity of light
$\lambda$ = wavelength of the light
${{v}_{o}}=\dfrac{c}{\lambda }=\dfrac{3.0\times {{10}^{8}}}{6800\times {{10}^{-10}}}=4.41\times {{10}^{14}}/s$
- From the threshold frequency we can calculate the work function and it is as follows.
Work function =

& h{{v}_{o}}=6.626\times {{10}^{-34}}\times 4.41\times {{10}^{14}} \\\ & =2.92\times {{10}^{-19}}J \\\ \end{aligned}$$ \- Therefore the Threshold frequency of the radiation is 4.41$\times {{10}^{14}}/s$. The work function for the radiation is 2.92$\times {{10}^{-19}}J$ . **Note:** First we have to calculate the threshold frequency of the radiation and later we have to calculate the work function with the help of the calculated threshold frequency. We cannot calculate the work function without threshold frequency.