Question

If the energy of incident photo and work function of metal are $E eV\text{ and }{{\phi }_{0}}eV$ respectively, then the maximum velocity of emitted photoelectrons will be –

& \text{A) }\dfrac{2}{m}(E-{{\phi }_{0}}) \\\ & \text{B) }\sqrt{\dfrac{2}{m}(E-{{\phi }_{0}})} \\\ & \text{C) }\sqrt{\dfrac{m}{2}(E-{{\phi }_{0}})} \\\ & \text{D) 2m}\sqrt{E-{{\phi }_{0}}} \\\ \end{aligned}$$

Answer 1

Photoelectric effect involves the emission of an electron due to the energy possessed by an electron, which hits a target with minimum kinetic energy. We can solve the velocity required for the electron to produce a photon from the kinetic energy.

Complete step-by-step solution
The photoelectric effect is the phenomenon in which a photon with the minimum energy releases an electron from a target metal which has a work function less than or energy of the photon. The photon under observation requires a threshold frequency to contribute to the energy requirement for ionizing the electron. The work function (${{\phi }_{0}}$) of the metal is the minimum energy required for releasing an electron from the metal.
Now, let us mathematically relate the above statements:
i.e., For a photoelectric effect to occur,

& \text{Energy of photon, E = Work function of the metal, }{{\phi }_{0}} \\\ & \Rightarrow \text{ }E={{\phi }_{0}} \\\ \end{aligned}$$ This is the minimum energy for the photoelectric effect in a given metal. Also, the greater energy can release an electron with the kinetic energy, so let us consider E as the energy possessed by the photon which is greater than the threshold energy or work function. The excess of energy is converted to the kinetic energy of the electron. i.e., $$E-\phi =\dfrac{1}{2}m{{v}^{2}}$$ So, we can derive the relation for velocity from the above equation as – $$v=\sqrt{\dfrac{2}{m}(E-{{\phi }_{0}})}$$ So, this is the maximum energy that we can observe in an electron released from a metal of work function $${{\phi }_{0}}$$ and a photon of energy E. **The correct answer is given by option B.** **Additional Information:** The photoelectric nature is the solid proof of the dual nature of electromagnetic waves. **Note:** We should understand the photons are packets of energy called quanta, they cannot add up to larger energy with a greater number of photons or with a higher intensity. i.e., Only one photon can interact with an electron at an instant. If the energy of this photon is less than the work function, then there is no photoelectric effect.