Question

An electron of mass 'm' and a photon have same energy 'E'. The ratio of de-Broglie wavelength of electron to the wavelength of photon is (c = velocity of light)

• A. $c\sqrt{\frac{E}{m}}$
• B. $\frac{1}{c}\sqrt{\frac{2m}{E}}$
• C. $\frac{1}{c}\sqrt{\frac{E}{2m}}$
• D. $c\sqrt{\frac{m}{E}}$

Hint:

You can verify the correct option quickly using dimensional analysis. Since a wavelength ratio is a dimensionless number, the right side of the equation must also be dimensionless. The term $\sqrt{\frac{E}{m}}$ has units of velocity ($\text{m s}^{-1}$). Dividing it by the speed of light $c$ ($\text{m s}^{-1}$) is the only way to cancel out all physical units.

Answer 1

The Correct Option is C

Step 1: Write the electron's de-Broglie wavelength.
A particle of mass $m$ with kinetic energy $E$ has momentum $p = \sqrt{2mE}$, so $$\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}.$$
Step 2: Write the photon's wavelength.
A photon of energy $E$ obeys $E = \dfrac{hc}{\lambda_p}$, giving $$\lambda_p = \frac{hc}{E}.$$
Step 3: Form the required ratio.
$$\frac{\lambda_e}{\lambda_p} = \frac{h/\sqrt{2mE}}{hc/E}.$$
Step 4: Cancel Planck's constant.
The $h$ cancels, leaving $$\frac{\lambda_e}{\lambda_p} = \frac{1}{\sqrt{2mE}}\times\frac{E}{c} = \frac{E}{c\sqrt{2mE}}.$$
Step 5: Take the energy inside the root.
Write $E = \sqrt{E^2}$ and combine under the radical: $$\frac{\lambda_e}{\lambda_p} = \frac{1}{c}\sqrt{\frac{E^2}{2mE}}.$$
Step 6: Simplify.
Cancelling one factor of $E$ inside the root, $$\frac{\lambda_e}{\lambda_p} = \frac{1}{c}\sqrt{\frac{E}{2m}}.$$
\[ \boxed{\frac{\lambda_e}{\lambda_p} = \frac{1}{c}\sqrt{\frac{E}{2m}}} \]