Question

Threshold wavelength for photoelectric effect on sodium is $$5000\stackrel{0}{A}$$. Its work function is
A. 15 J
B. $$16\times {10}^{-14}J$$
C. $$4\times {10}^{-19}J$$
D. $$4\times {10}^{-18}J$$

Answer 1

Hint:Work function is defined as the minimum energy which is required by an electron to escape from the metal surface. On the other hand, we say that the work function is the energy that is needed to eject electrons from the metal surface. So, the maximum energy (E) of the photoelectron can be equal to the energy of the incident photon (h) minus the work function ($$\varphi$$), because an electron has to do some kind of work to escape the potential from the metal surface.

Formula used
Work function is given as:
$$\varphi =h{\upsilon }_0={\displaystyle \frac{hc}{{\lambda }_0}}$$
Where, h is Planck's constant, $${\upsilon }_0$$ is the threshold frequency, c is the speed of light and $${\lambda }_0$$ is the wavelength.

Complete step by step solution:
Given threshold wavelength for photoelectric effect on sodium,
$$\lambda =5000\stackrel{0}{A}$$
As we know that work function,
$$\varphi ={\displaystyle \frac{hc}{{\lambda }_0}}$$
$$\Rightarrow \varphi ={\displaystyle \frac{6.625\times {10}^{-34}}{5000\times {10}^{-10}}}$$
$$\therefore \varphi =4\times {10}^{-19}J$$

Hence option C is the correct answer

Additional information: The photoelectric effect is defined as the phenomenon in which electrically charged particles are released from a material after it absorbs electromagnetic radiation. This photoelectric effect is also defined as the emission of electrons from a metal surface when light falls on it.

Note: In a metal surface, the threshold frequency is defined as the minimum frequency of incident radiation which is required for the photoelectric effect to happen. When the incident radiation has a frequency below the threshold frequency of the metal surface then there will be no emission of photoelectrons. It is usually represented $$\lambda or{\lambda }_0$$.