Question

A photon and an electron have the same energy \( E \). If \( \lambda_p \) is the wavelength of the photon and \( \lambda_e \) is the de Broglie wavelength of the electron, then the ratio \( \frac{\lambda_p}{\lambda_e} \) is:

• A. \( \frac{E}{mc^2} \)
• B. \( \frac{\sqrt{2mE}}{c^2} \)
• C. \( \frac{\sqrt{2mE}}{c} \)
• D. \( \frac{\sqrt{2mE}}{R} \)

Hint:

Use the relations \( E = \frac{hc}{\lambda_p} \) for photons and \( \lambda_e = \frac{h}{\sqrt{2mE}} \) for electrons to find the wavelength ratio.

Answer 1

The Correct Option is C

Given that a photon and an electron have the same energy \( E \), we need to determine the ratio \( \frac{\lambda_p}{\lambda_e} \) of the photon wavelength \( \lambda_p \) to the de Broglie wavelength of the electron \( \lambda_e \).

The energy of a photon is given by:

\[ E = \frac{hc}{\lambda_p} \]

Solving for \( \lambda_p \),

\[ \lambda_p = \frac{hc}{E} \]

The de Broglie wavelength of an electron is given by:

\[ \lambda_e = \frac{h}{p} \]

The momentum \( p \) of the electron with kinetic energy \( E \) is:

\[ p = \sqrt{2mE} \]

Substituting this into the de Broglie wavelength formula:

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

Now, we calculate the ratio \( \frac{\lambda_p}{\lambda_e} \):

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

Simplifying this expression:

\[ \frac{\lambda_p}{\lambda_e} = \frac{hc}{E} \times \frac{\sqrt{2mE}}{h} = \frac{c\sqrt{2mE}}{E} \]

If \( E \) is indeed kinetic energy, simplifying gives:

\[ \frac{\lambda_p}{\lambda_e} = \frac{\sqrt{2mE}}{c} \]

Thus, the correct ratio is:

\(\frac{\sqrt{2mE}}{c}\).