MHTCET 19 Apr 2026 Shift 2 Paper

Concept:Einstein’s photoelectric equation relates the stopping potential to the incident wavelength and the threshold wavelength.Explanation:For wavelength $\lambda$, the stopping potential is $V$:$$eV = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)$$For wavelength $1.5\lambda$, the stopping potential is $\frac{V}{4}$:$$\frac{eV}{4} = hc\left(\frac{1}{1.5\lambda} - \frac{1}{\lambda_0}\right) = hc\left(\frac{2}{3\lambda} - \frac{1}{\lambda_0}\right)$$Divide the first equation by $4$ and equate it with the second equation:$$\frac{1}{4}\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right) = \frac{2}{3\lambda} - \frac{1}{\lambda_0}$$Multiplying throughout by $12$:$$\frac{3}{\lambda} - \frac{3}{\lambda_0} = \frac{8}{\lambda} - \frac{12}{\lambda_0}$$Rearranging terms gives:$$\frac{9}{\lambda_0} = \frac{5}{\lambda}$$Hence:$$\lambda_0 = \frac{9}{5}\lambda$$Answer:The threshold wavelength is $\frac{9}{5}\lambda$, so the correct option is D.