Question

The photoelectric threshold for a metal surface is $$6600\dot{\mathrm{A}}$$. The work function for this is:
A. $$1.87V$$
B. $$1.87eV$$
C. $$18.7eV$$
D. $$0.18eV$$

Answer 1

Hint:n photoelectric effect, the minimum amount of energy required for electron emission from the metal surface is the work function of that metal and the frequency of light corresponding to this minimum energy is called threshold frequency and the corresponding wavelength is called threshold wavelength. It is calculated as work function, $${\varphi }_o=h{\nu }_{threshold}={\displaystyle \frac{hc}{{\lambda }_{threshold}}}$$.

Formula Used:
Energy, $$E=h\nu ={\displaystyle \frac{hc}{\lambda }}$$
Equation for photoelectric effect, $$h\nu ={\varphi }_o+E_k$$
Work function, $${\varphi }_o=h{\nu }_{threshold}={\displaystyle \frac{hc}{{\lambda }_{threshold}}}$$
Where,
E = incident energy
$${\varphi }_o$$= Work function of the metal
$$E_k$$= energy of the emitted photoelectron
$$c$$ = speed of light = $$3\times {10}^{8}m/s$$
$$\nu$$= frequency of the light,
$${\nu }_{threshold}$$= threshold frequency,
$$\lambda$$= wavelength of the light,
$${\lambda }_{threshold}$$= threshold wavelength

Complete step by step solution:
Photoelectric effect was discovered by Einstein in 1905 for which he also won the Nobel prize in physics. According to its theory, when a light of sufficient energy is incident on a metal surface, the photons of the incident light impart energy to the electrons on the metallic surface. If this energy is higher than the threshold energy of the metal, the electrons become sufficiently energetic to escape the metal surface and are emitted. These electrons are called photoelectrons whose energy is less than the energy of the incident light as some of the energy is utilised in overcoming the barrier energy or the work function.

Given: Threshold wavelength for a particular metallic surface , $${\lambda }_{threshold}$$ = $$6600\dot{\mathrm{A}}$$. $$c=speedoflight=3\times {10}^{8}m/s$$
We need to determine the work function.
Equation for the photoelectric effect is $$h\nu ={\varphi }_o+E_k$$. Here we have to calculate the work function which is given by :
$${\varphi }_o=h{\nu }_{threshold}={\displaystyle \frac{hc}{{\lambda }_{threshold}}}$$
$$\Rightarrow {\varphi }_o={\displaystyle \frac{hc}{{\lambda }_{threshold}}}$$
We know, $$1\dot{A}={10}^{-10}m$$
$${\varphi }_o={\displaystyle \frac{6.64\times {10}^{-34}\times 3\times {10}^8}{6600\times {10}^{-10}}}$$
$$\Rightarrow {\varphi }_o=3\times {10}^{-19}J$$
We need the answer in eV. We know that $$1eV=1.6\times {10}^{-19}J$$
$${\varphi }_o={\displaystyle \frac{3\times {10}^{-19}}{1.6\times {10}^{-19}}}$$ eV
$$\therefore {\varphi }_o=1.87eV$$

Hence option B is the correct answer.

Note: No emission of electrons or any photoelectric effect will take place if the energy supplied is less than the work function. Work function is a property of the metal that depends on the metal. For e.g., work function is different for different metals like Gold, Silver among others. It is independent of the nature or the property of the incident light. From the total energy supplied on the metal surface , the remaining energy after the work function is responsible for emission of the electron from the surface. This extra energy is converted to kinetic energy which enables the electron to emit from the metal surface.