Question

An electron and a photon have the same wavelength of ${10^{ - 9}}\,{\text{m}}$. If E is the energy of the photon and p is the momentum of the electron, the magnitude of $E/P$ in SI units is?
(A) $1.00 \times {10^{ - 9}}$
(B) $1.50 \times {10^8}$
(C) $3.00 \times {10^8}$
(D) $1.20 \times {10^7}$

Answer 1

Initially you must be aware of the general formula of momentum and photon. De-Broglie wavelength plays an important factor in this question. Deriving a relation between energy and momentum by wavelength will surely help you.

Complete step by step solution:
Given:
$\lambda = {10^{ - 9}}\;{\text{m}}$
Formula:
$E = \dfrac{{hc}}{\lambda }$
Calculation:
From Planck's equation
$\Rightarrow$ $E = \dfrac{{hc}}{\lambda }$
$\Rightarrow$ ${\lambda _p} = \dfrac{{hc}}{E}$​
From de-Broglie’s equation
$\Rightarrow$ ${\lambda _e} = \dfrac{h}{p}$
Since,
$\Rightarrow$ ${\lambda _p} = {\lambda _e}$
$\Rightarrow$ $\dfrac{{hc}}{E} = \dfrac{h}{p}$
$\Rightarrow$ $\dfrac{E}{p} = c$
$= 3 \times {10^8}$

Therefore, option C is a correct option.

Note:
DE Broglie wavelength plays an important role in physics. It is said that matter incorporates a dual nature of wave-particles. Nuclear physicist waves, named after the discoverer Louis nuclear physicist, is that the property of a fabric object that varies in time or space while behaving the same as waves. It's also called matter-waves. It holds great similarity to the twin nature of sunshine which behaves as particle and wave, which has been proven experimentally.