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# The work function for a certain metal is $4.2eV$. Will this metal give photoelectric emission for incident radiation of wavelength $330nm$?

Last updated date: 13th Jun 2024
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Hint: The process in which photoelectrons are ejected from the surface of the metal due to the action of light is known as photoelectric emission or photoemission. The photoelectric emission only takes place when the energy of incident radiation is more than the work function of a metal.

As we know, the work function of a metal is the minimum energy required by an electron to just escape from the metal surface so as to overcome the restraining forces at the surface. Work function of a metal is generally denoted by $\Phi^{o}$ and it is usually measured in $eV$ (electron Volt). The phenomenon of emission of electrons from the surface of a metal is called electron emission. The electrons in the metal at room temperature can not leave the metal surface due to the restraining surface and the energy possessed by the electron is less than the work function of the metal. Therefore, the electron emission from a metal surface is possible if the energy greater than the work function of metal is supplied to the metal. The emission of electron can be done by following process –
(i) Thermionic process: - It is the phenomenon of emission of electrons from metal surfaces when heated suitably.
(ii) Photoelectric emission: - It is the emission of electrons from the surface of metal when light radiation of suitable frequency falls on it.
(iii) Field emission or cold cathode emission: - It is the phenomenon of emission of electrons from the surface of a metal under the application of a strong electric field.
(iv) Secondary emission: - It is a phenomenon of emission of electrons from the surface of metal in large numbers when fast moving electrons strike the metal surface.
Now, according to question, the given work function of metal is $4.2eV$. The energy of incident radiation can be calculated by using: -
$E = \dfrac{{hc}}{{\lambda \times 1.6 \times {{10}^{ - 19}}}}eV$……… (i) $\left[ {\because E = \dfrac{{hc}}{\lambda }J} \right]$
Here, $h$ is Planck’s constant $6.63 \times {10^{ - 34}}Js$, $c$ is speed of light (in vaccum) i.e. $3 \times {10^8}m/s$ and $\lambda$ is the wavelength of light radiation whose value given is $330nm$ or $330 \times {10^{ - 9}}m$ [because $1mn = 1.6 \times {10^{ - 9}}m$]
Putting the values in equation (i), we get
$\Rightarrow$ $E = \dfrac{{6.63 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}m{s^{ - 1}}}}{{330 \times {{10}^{ - 9}}m \times 1.6 \times {{10}^{ - 19}}}} = 3.7eV$
Since energy of radiation is less than the work function of a metal surface, the photoelectric effect is not shown by this metal surface if a radiation of wavelength $330nm$ incident on it.

Note: We have an energy equation normally as $E = \dfrac{{hc}}{\lambda }$. However in equation (i) we have divided the equation with $1.6 \times 1{0^{ - 19}}$ to obtain the value in the $eV$ whereas, from normal equation, we get the value of energy in Joule, $1eV = 1.6 \times {10^{ - 19}}J$. Moreover, the minimum amount of energy that is required to remove an electron from the metal surface is known as threshold energy.