MHTCET 17 Apr 2026 Shift 1 Paper

Concept:The de Broglie wavelength is inversely proportional to the square root of the kinetic energy of the electron.Explanation:The de Broglie wavelength of a free electron is given by $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the linear momentum.The kinetic energy is expressed as $K = \frac{p^2}{2m}$, which gives $p = \sqrt{2mK}$.Substituting this value of $p$ into the wavelength formula, we get $\lambda = \frac{h}{\sqrt{2mK}}$.This shows that $\lambda \propto \frac{1}{\sqrt{K}}$, so the wavelength is inversely proportional to the square root of kinetic energy.When the kinetic energy becomes three times the original value, $K' = 3K$.The ratio of the new de Broglie wavelength $\lambda'$ to the previous wavelength $\lambda$ is:$$\frac{\lambda'}{\lambda} = \sqrt{\frac{K}{K'}} = \sqrt{\frac{K}{3K}} = \frac{1}{\sqrt{3}}$$Answer:Therefore, the ratio of the new de Broglie wavelength to the previous de Broglie wavelength is $\frac{1}{\sqrt{3}}$, which corresponds to Option A.