Question

The wavelength '\(\lambda\)' of photon and the de-Broglie wavelength of an electron have same value. The ratio of kinetic energy of electron to the energy of a photon is (m=mass of electron, h = Planck's constant, c = velocity of light)

• A. \(\frac{h}{2\lambda mc}\)
• B. \(\frac{h\lambda }{2mc}\)
• C. \(\frac{2c}{hm\lambda }\)
• D. \(\frac{m\lambda }{2hc}\)

Hint:

Express the electron kinetic energy through its de Broglie momentum and divide it by the photon energy hc over lambda.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Photon energy is \(E=\dfrac{hc}{\lambda}\). An electron with de Broglie wavelength \(\lambda\) has momentum \(p=\dfrac{h}{\lambda}\).

Step 2: Kinetic energy of electron
\(K=\dfrac{p^2}{2m}=\dfrac{h^2}{2m\lambda^2}\).

Step 3: Ratio
\(\dfrac{K}{E}=\dfrac{h^2/(2m\lambda^2)}{hc/\lambda}=\dfrac{h}{2\lambda mc}\).

Final Answer:
The ratio of electron kinetic energy to photon energy is \(\dfrac{h}{2\lambda mc}\). \[ \boxed{\text{(A) }\dfrac{h}{2\lambda mc}} \]