Question

An electron of mass $m$ and a photon have the same energy $E$ . The ratio of de-Broglie wavelength associated with them is:
A. $\dfrac{1}{c}{\left( {\dfrac{E}{{2m}}} \right)^{\dfrac{1}{2}}}$
B. ${\left( {\dfrac{E}{{2m}}} \right)^{\dfrac{1}{2}}}$
C. $c{\left( {2mE} \right)^{\dfrac{1}{2}}}$
D. $\dfrac{1}{c}{\left( {\dfrac{{2m}}{E}} \right)^{\dfrac{1}{2}}}$ (c being velocity of light)

Answer 1

To solve this question, we must have the knowledge of de-Broglie’s equation.
$\lambda = \dfrac{h}{p}$ where, $h$ is the planck's constant and $p$ is the linear momentum.
Change $p$ in terms of energy $E$ .

Complete step-by-step solution:
According to de Broglie's hypothesis of matter waves postulates that any particle of matter that has linear momentum is also a wave. The wavelength of a matter wave associated with a particle is inversely proportional to the magnitude of the particle's linear momentum.
Since we already know the de-Broglie’s equation $\lambda = \dfrac{h}{p} - - - (1)$ and for a particle $p = \sqrt {2mE}$ .
Where,$h$ is the planck's constant,$p$ is the linear momentum, $\lambda$ is the de-Broglie wavelength, $E$ is the energy.
For electron of mass $m$, according to de-Broglie’s equation as per equation (1)
So ${\lambda _1} = \dfrac{h}{{\sqrt {2mE} }} - - - (2)$
This was the case for electrons but what happens in case of photons let’s see.
We know that for photon’s rest mass is zero and for zero rest mass particle we can use relation that momentum is linear variance of energy so, for photon we can say
$p = \dfrac{E}{c}$
Putting this in de-Broglie’s equation as per equation (1), we get
${\lambda _2} = \dfrac{{hc}}{E} - - - (3)$
On further as per question taking ratio of equation (2) and (3)
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{1}{c}\sqrt {\dfrac{E}{{2m}}}$
Further simplifying
$\Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{1}{c}{\left( {\dfrac{E}{{2m}}} \right)^{\dfrac{1}{2}}}$

Note:- For photons the energy depends on frequency as $E \propto \nu$ where $\nu$ is the frequency of the photon. So, total energy $E$ associated with photons is $E = nh\upsilon$ where, $h$ is the planck's constant, $E$ is the total energy, $n$ is the total number of photons and $\nu$ is the frequency of the photon.