Question

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

• A. $\bigg( \frac{E}{2m} \bigg)^{\frac{1}{2}}$
• B. $c(2mE )^{\frac{1}{2}}$
• C. $\frac{1}{c} \bigg( \frac{2m}{E} \bigg)^{\frac{1}{2}}$c
• D. $\frac{1}{c} \bigg( \frac{E}{2m} \bigg)^{\frac{1}{2}}$c

Answer 1

The Correct Option is D

To solve this problem, we will determine the de-Broglie wavelengths for both an electron and a photon with the same energy and then find the ratio of these wavelengths.

Step 1: Understand the de-Broglie Wavelength

The de-Broglie wavelength (\lambda) is given by the formula:

\lambda = \frac{h}{p}

where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

Step 2: Calculate the Wavelength for an Electron

For an electron, given its energy \( E \), the relation between energy and momentum is:

E = \frac{p^2}{2m}

Rearranging to find momentum \( p \), we have:

p = \sqrt{2mE}

Substituting this in the de-Broglie wavelength formula:

\lambda_e = \frac{h}{\sqrt{2mE}}

Step 3: Calculate the Wavelength for a Photon

For a photon, the energy is related to momentum by:

E = pc

Solving for momentum \( p \) gives:

p = \frac{E}{c}

Substitute back into the de-Broglie formula:

\lambda_p = \frac{h}{\frac{E}{c}} = \frac{hc}{E}

Step 4: Find the Ratio of Wavelengths \frac{\lambda_e}{\lambda_p}

The ratio of the electron's de-Broglie wavelength to the photon's de-Broglie wavelength is:

\frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}}

Simplifying this expression:

\frac{\lambda_e}{\lambda_p} = \frac{E}{c\sqrt{2mE}}

Further simplification gives us:

\frac{\lambda_e}{\lambda_p} = \frac{1}{c}\sqrt{\frac{E}{2m}}

Conclusion

The ratio of the de-Broglie wavelengths of an electron and a photon with the same energy is \frac{1}{c} \bigg( \frac{E}{2m} \bigg)^{\frac{1}{2}}. This matches option c.