Question

The work function of tungsten is 4.50eV. The wavelength of the fastest electron emitted when light whose photon energy is 5.50 eV falls on a tungsten surface, is?
(A) 12.27
(B) 0.286
(C) 12400
(D) 1.227

Answer 1

The work function is a term used in photoelectric effect which describes the energy required for an electron to bring it from inside of a metal to its surface. If the light incident has more energy than the work function, then the electron comes out of the metal along with an additional kinetic energy.
Formula used:
The relation for work function/threshold energy in a metal is given as:
$\dfrac{hc}{\lambda} = \dfrac{hc}{\lambda'} + \phi_0$
where primed variable (') is for the additional kinetic energy of the emitted electron.

Complete answer:
We are given that the work function of the tungsten metal is 4.50 eV, the incident light has the energy of 5.50 eV. The incident energy is used in providing the threshold energy and the difference of incident and threshold will give us the energy of the fastest electron emitted:
$\dfrac{hc}{\lambda'} = 5.50 - 4.50 = 1.0$ eV.
We can directly substitute the value of 'hc' here as:
$hc = 12400 eV \buildrel _\circ \over {\mathrm{A}}$.
We substitute the values and get:
$\dfrac{12400 \text{eV} \buildrel _\circ \over {\mathrm{A}} }{1 eV} = \lambda' = 12400 \buildrel _\circ \over {\mathrm{A}}$ .
The electron comes out with a wavelength of 12400 $\buildrel _\circ \over {\mathrm{A}}$.

Therefore, the correct option is (C).

Additional Information:
The photoelectric effect constitutes emission of electrons when a light of certain energy above threshold of the metal of the cathode is incident on the cathode. Cathode being negatively charged and anode being positively charged gives rise to an arrangement that helps in the flow of the emitted electron.

Note:
One should take note of the units very carefully while handling such problems. All the given energies are in electron volts. The product of Planck’s constant and speed of light is written in terms of electron volts ang angstroms to facilitate the calculations.