MHTCET 16 Apr 2026 Shift 2 Paper

Concept:Einstein's photoelectric equation relates incident photon energy, the work function of the metal, and the maximum kinetic energy of emitted photoelectrons.Explanation:For the initial wavelength $\lambda$, the maximum kinetic energy of the emitted photoelectrons is given by:$$\frac{1}{2}mV^{2} = \frac{hc}{\lambda} - \phi$$Here, $\phi$ is the work function of the photosensitive surface.When the wavelength is changed to $\lambda' = \frac{2\lambda}{3}$, the new photon energy becomes:$$\frac{hc}{\lambda'} = \frac{hc}{2\lambda/3} = \frac{3hc}{2\lambda} = 1.5\left(\frac{hc}{\lambda}\right)$$Let the new maximum speed be $V'$. Using the photoelectric equation again:$$\frac{1}{2}mV'^{2} = \frac{3hc}{2\lambda} - \phi$$Substituting $hc/\lambda = \phi + \frac{1}{2}mV^{2}$ into the above equation:$$\frac{1}{2}mV'^{2} = 1.5\left(\phi + \frac{1}{2}mV^{2}\right) - \phi$$Simplifying:$$\frac{1}{2}mV'^{2} = 1.5\left(\frac{1}{2}mV^{2}\right) + 0.5\phi$$Since the work function $\phi$ is always positive, the additional term $0.5\phi$ increases the kinetic energy.Therefore:$$\frac{1}{2}mV'^{2} > 1.5\left(\frac{1}{2}mV^{2}\right)$$$$\Rightarrow V'^{2} > 1.5V^{2}$$$$\Rightarrow V' > V(1.5)^{\frac{1}{2}}$$Answer:The maximum speed of emitted photoelectrons will be greater than $V(1.5)^{\frac{1}{2}}$.Correct option: B.