Kerala KEAM 5 June 2024 Solved Paper

We know that the kinetic energy of the emitted photoelectrons is given by the photoelectric equation: $$K.E. = h f - \phi$$where:- $K.E.$ is the kinetic energy of the emitted photoelectrons,- $h$ is Planck's constant,- $f$ is the frequency of the incident light,- $\phi$ is the work function of the metal.Let the initial frequency be $f_1$ and the corresponding kinetic energy be $K.E_1 = 0.5 \, eV$: $$K.E_1 = h f_1 - \phi \tag{1}$$Let the final frequency be $f_2 = 1.1 f_1$ (since the frequency is increased by 10\%), and the corresponding kinetic energy be $K.E_2 = 0.7 \, eV$: $$K.E_2 = h f_2 - \phi = h (1.1 f_1) - \phi \tag{2}$$Now, subtract equation (1) from equation (2): $$K.E_2 - K.E_1 = h (1.1 f_1) - \phi - (h f_1 - \phi)$$ $$0.7 - 0.5 = h (1.1 f_1 - f_1)$$ $$0.2 = h \times 0.1 f_1$$ $$h f_1 = 2 \, eV$$Substitute this value into equation (1): $$0.5 = 2 - \phi$$ $$\phi = 1.5 \, eV$$Thus, the work function of the metal is 1.5 eV, which corresponds to option (C). Quick Tip: In photoelectric effect problems, the change in kinetic energy of photoelectrons can be used to find the work function by applying the photoelectric equation and solving for $\phi$.