Question

Photoelectric work function of a metal is $1eV$. Light of wavelength $300nm$ falls on it. The photoelectron come out with velocity
A. $10m{s}^{-1}$
B. $100m{s}^{-1}$
C. $10000m{s}^{-1}$
D. $1000000m{s}^{-1}$

Answer 1

Hint:Here, metal is exposed to light and you are asked to find the velocity with which the electrons will come out. Here, the particle behaviour of light is to be considered as you know that light has dual nature. So, photoelectric effect is a phenomenon in which when metal is exposed to light, due to the energy provided by light to the metal, the electrons of metal will leave the metal surface and come out. It is not like you give light and the metal will throw out the electrons, which is why the term work function is defined.

Complete step by step answer:
When light of sufficiently small wavelength, that is high energy, is incident on a metal surface, the electrons of the metal are ejected and this phenomenon is called photoelectric effect. Let us denote the energy of the light incident on the metal surface by $E$, this much amount of energy is given to the metal and in turn, electrons are ejected. The energy required to just eject the electron is called the work function of that metal and is denoted by $\phi$.

The remaining part of the energy $E$ is given to the electrons so that they can move, which means that the remaining part is converted into the kinetic energy of the electrons. You can see that $E=\phi +K_{max}$, where $$K_{max}$$ is the maximum kinetic energy that can be achieved by the electron and this maximum value can only be achieved if the electron does not make any collisions while escaping from the metal surface.So, we have
$$\begin{gathered} K_{max}=E-\phi \\ \Rightarrow {\displaystyle \frac{1}{2}}m{v}^2=E-\phi \\ \Rightarrow v=\sqrt{{\displaystyle \frac{2\left(E-\phi \right)}{m}}} \end{gathered}$$
Energy of light can be given by the equation $E={\displaystyle \frac{hc}{\lambda }}$, therefore, we get $v=\sqrt{{\displaystyle \frac{2\left({\displaystyle \frac{hc}{\lambda }}-\phi \right)}{m}}}$.
Let us substitute the values we have.
$$\begin{gathered} v=\sqrt{{\displaystyle \frac{2\left({\displaystyle \frac{hc}{\lambda }}-\phi \right)}{m}}} \\ \Rightarrow v=\sqrt{{\displaystyle \frac{2\left({\displaystyle \frac{\left(6.626\times {10}^{-34}\right)\left(3\times {10}^8\right)}{\left(300\times {10}^{-9}\right)}}-1\times 1.6\times {10}^{-19}\right)}{\left(9.1\times {10}^{-31}\right)}}} \\ \Rightarrow v=\sqrt{1.1046\times {10}^{12}} \\ \therefore v\approx {10}^{6}m{s}^{-1} \end{gathered}$$
Therefore, the photoelectron comes out with velocity $1000000m{s}^{-1}$ and thus option D is correct.

Note:Keep in mind the phenomenon of the photoelectric effect. Also remember that it is not for sure that the electron will come out, we consider it as, if an electron comes out without any loss of energy, then it will have kinetic energy equal to $K_{max}$. Also, keep in mind that the equation which we used to calculate the velocity of the electron coming out is known as Einstein's photoelectric equation.