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

Electron and Photon Wavelength Comparison 

For an electron, the de Broglie wavelength is given by:

\(\lambda_e = \frac{h}{\sqrt{2mE}}\),

where:

• h is Planck's constant
• m is the mass of the electron
• E is the energy of the electron

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

\(E = pc \, \Rightarrow \, \lambda_{Ph} = \frac{hc}{E}\),

where:

• p is the momentum of the photon
• c is the speed of light
• E is the energy of the photon

Now, comparing the wavelengths of the electron and the photon:

\(\Rightarrow \frac{\lambda_e}{\lambda_{Ph}} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \left( \frac{E}{2m} \right)^{1/2} \frac{1}{c}\)

Conclusion:

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

\(\frac{\lambda_e}{\lambda_{Ph}} = \left( \frac{E}{2m} \right)^{1/2} \frac{1}{c}\).