Question

Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength ${6800}^{°}A$.Calculate threshold frequency $v_{\circ }$ and work function $w_{\circ }$ of the metal.

Answer 1

Hint: To solve this, firstly find out the threshold frequency. You can use the formula, $\upsilon ={\displaystyle \frac{c}{{\lambda }_{\circ }}}$. Then find out the work function from here using the formula $\varphi =h{\upsilon }_o$

Complete answer:
First let us know about a few terms which we will be using to solve the given question!
1. Kinetic energy:
Kinetic energy is the energy associated with the movement of objects. The kinetic energy of an object depends on both its mass and velocity, with its velocity playing a much greater role. Example of kinetic energy, an airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity.
Kinetic energy (E) of a massive body in motion can be calculated as half its mass (m) times the square of its velocity (v):
 $E={\displaystyle \frac{1}{2}}{mv}^2$
2. Work Function:
It is the minimum quantity of energy which is required to remove an electron to infinity from the surface of a given solid, usually a metal. The work function is one of the fundamental electronic properties of bare and coated metallic surfaces.
Now, let us discuss about the work function equation:
The photoelectric equation involves; h = the Planck constant and its value is $6.63\times {10}^{-34}$ J s.
Frequency of the incident light in hertz = the work function in joules (J). $E_k$ is the maximum kinetic energy of the emitted electrons in joule (J).
3. Threshold frequency:
It is the frequency of light that carries enough energy to dislodge an electron from an atom. This energy is entirely consumed in the process. Therefore, the electron gets no kinetic energy at the threshold frequency and it is not released from the atom.
Now, let us try to solve the given question!
Given, the wavelength of radiation is: ${6800}^{°}A$
Energy given = Work function + Kinetic energy
Since, the electrons are emitted with zero velocity from a metal surface when it is exposed to radiation. Which means the kinetic energy will be zero.
Threshold frequency will be, $\upsilon ={\displaystyle \frac{c}{{\lambda }_{\circ }}}={\displaystyle \frac{3.0\times {10}^{8}m/s}{6800\times {10}^{-19}m}}=4.14\times {10}^{14}$
So, the work function will be:
$h{\upsilon }_o=6.626\times {10}^{-34}\times 4.41\times {10}^{14}=2.92\times {10}^{-19}$ J and this is the required answer.

Note:
The photoelectric effect was discovered in 1887 by the German physicist Heinrich Rudolf Hertz. He observed that when ultraviolet light shines on two metal electrodes with a voltage applied across them, the light changes the voltage at which sparking takes place.
The photoelectric effect was explained by Einstein.
He explained that the electrons were emitted from the metal surface when the light of sufficient frequency is incident which implies kinetic energy of electrons increases with light intensity.