Kelvi

An electron of mass m and a photon have the same energy E. The ratio of de Broglie wavelengths associated with them is

medium

Dual Nature of Radiation and Matter

2016

physics

$c(2mE)^{1/2}$

$\frac{1}{c} \left( \frac{2m}{E} \right)^{1/2}$

$\frac{1}{c} \left( \frac{E}{2m} \right)^{1/2}$

$\left( \frac{E}{2m} \right)^{1/2}$

Explanation

To solve this problem, we need to find the de Broglie wavelengths for both the electron and the photon and then find their ratio. $\\$ The de Broglie wavelength $\lambda$ is given by: $\\ \lambda = \frac{h}{p} \\$ where $h$

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