If the de Broglie wavelength of a moving electron is numerically equal to exactly half the distance

Question

If the de Broglie wavelength of a moving electron is numerically equal to exactly half the distance travelled by it in one second, then the correct mathematical relation is

Professor PC Thomas Classes · Answer

(a) : The magnitude of the distance travelled by the electron in one second is numerically identical to its velocity $v$. Given that the wavelength is $\lambda = v/2$, we have $v = 2\lambda$. According to the de Broglie relation, $\lambda = \frac{h}{mv}$. Substituting $v = 2\lambda$ into this equation gives $\lambda = \frac{h}{m(2\lambda)}$, which implies $2\lambda^2 = \frac{h}{m}$. Solving for wavelength gives $\lambda = \sqrt{\frac{h}{2m}}$.