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Question: Name the unit in which the work function of a metal is generally expressed. How is it related to the...

Name the unit in which the work function of a metal is generally expressed. How is it related to the SI unit joule?
A. eV, 1eV=1.6×109J B. eV, 1eV=1.6×10−9J C. eV, 1eV=1.6×10−19J D. eV, 1eV=1.6×1019J \begin{aligned} & \text{A}\text{. }eV,\text{ }1eV=1.6\times {{10}^{9}}J \\\ & \text{B}\text{. }eV,\text{ }1eV=1.6\times {{10}^{-9}}J \\\ & \text{C}\text{. }eV,\text{ }1eV=1.6\times {{10}^{-19}}J \\\ & \text{D}\text{. }eV,\text{ }1eV=1.6\times {{10}^{19}}J \\\ \end{aligned}

Explanation

Solution

The work function of a metal is the minimum amount of energy required to knock out the electron from the metal surface when the photon of sufficient energy falls on it. Here, the energy of the light is transferred to an electron to escape out. Thus, the work function is expressed in the same unit as that of the energy of the photon.

Complete step by step answer:
When a photon of sufficient energy falls on the metal surface it knocks out the outermost electrons from the metal. This phenomenon is called the photoelectric effect. The knocked electrons are called ‘Photoelectrons’.
The minimum energy required to knock out electrons from the metal surface is called the work function (φ\varphi ) of that metal. If the photon has energy more than the work function of metal falls on it then the rest of the energy is converted into kinetic energy of the electron.
Therefore, when a photon of a certain frequency ν\nu on a cathode having work function φ\varphi then the kinetic energy of the ejected electron is given by Einstein’s photoelectric equation.
K.E.=hν−φK.E.=h\nu -\varphi
Where h is Planck’s constant and hνh\nu is the energy of the incident light.
Here, we are dealing with electrons and the energy required to knock out and attract electrons. Hence, it is very convenient to express energy in terms of energy of electrons (charged particles) moving in a potential difference.
We define the amount of energy lost or gained by an electron while moving through a potential difference of 1V as 1eV.
So in the photoelectric effect as the kinetic energy and the energy of photons are expressed in terms of eV, we also express the work function φ\varphi of the metal in eV.
For any charge q moving in a potential difference of V volts, the energy gained or lost by is given as
E=q×V JE=q\times V\text{ }J
Since the charge on an electron is 1.6×10−19C1.6\times {{10}^{-19}}C, the energy lost or gained by an electron in the potential difference of 1V is
E=1.6×10−19×1 J=1.6×10−19JE=1.6\times {{10}^{-19}}\times 1\text{ }J=1.6\times {{10}^{-19}}J
Since this energy is also equal to 1eV, we can write,
1eV=1.6×10−19J1eV=1.6\times {{10}^{-19}}J
This is how 1eV is related to SI unit joule.

Answer - C. eV, 1eV=1.6×10−19J\text{C}\text{. }eV,\text{ }1eV=1.6\times {{10}^{-19}}J

Note:
An eV is a very convenient unit when we deal with the energy associated with electrons. This unit is commonly used in atomic and nuclear physics. In astronomy, the energy difference between spectral lines is expressed in terms of eV. You will find that MeV, GeV, and even TeV are commonly used in many fields.