If the kinetic energy of a moving proton is artificially increased to times its initial value, its n — Atomic Structure Chemistry Question

💡 Solution & Explanation

The de Broglie wavelength is related to kinetic energy by $\lambda = \frac{h}{\sqrt{2m(K.E.)}}$. Thus, $\lambda \propto \frac{1}{\sqrt{K.E.}}$. If the new kinetic energy $K.E.' = 9 \times K.E.$, the new wavelength $\lambda' \propto \frac{1}{\sqrt{9(K.E.)}} = \frac{1}{3\sqrt{K.E.}}$. Therefore, $\lambda' = \frac{\lambda}{3}$. Since it becomes $\frac{1}{k}$ times the original, $k = 3$.