Which of the following mathematical expressions correctly represent the de Broglie wavelength of a m — Atomic Structure Chemistry Question

💡 Solution & Explanation

The de Broglie wavelength is $\lambda = \frac{h}{p}$. Since momentum $p = mv$ and $K.E. = \frac{1}{2}mv^2$, we can write $p = \sqrt{2m(K.E.)}$, leading to $\lambda = \frac{h}{\sqrt{2m(K.E.)}}$. For a charged particle accelerated by potential $V$, $K.E. = qV$, giving $\lambda = \frac{h}{\sqrt{2mqV}}$. Option D has an incorrect square root on velocity.