Question

Let E_e and E_p represent the kinetic energy of electron and photon, respectively. If the de-Broglie wavelength λ_p of a photon is twice the de-Broglie wavelength λ_e of an electron, then `E_p/E_e` is ______.

(speed of electron = `c/100`, c = velocity of light)

Options

• 10^-1

• 10²

• 10^-2

• 10⁴

Answer 1

Let E_e and E_p represent the kinetic energy of electron and photon, respectively. If the de-Broglie wavelength λ_p of a photon is twice the de-Broglie wavelength λ_e of an electron, then `E_p/E_e` is 10².

Explanation:

For electron, `lambda_e = h/sqrt(2mE_e) ⇒ E_e = h^2/((2m)lambda_e^2)`

For photon, `lambda_p = (hc)/(E_p)`

⇒ `E_p = (hc)/lambda_p = (hc)/(2lambda_e)`  `(∵ lambda_p = 2lambda_e)`

⇒ `E_p/E_c = (hc)/(2lambda_e) xx (2mlambda_e^2)/h^2 = mc xx lambda_e/h`

Also, `lambda_e = h/(m"v"_e) ⇒ lambda_e/h = 1/(m"v"_e)`

⇒ `E_p/E_e = mc xx 1/(m"v"_c) = c/"v"_e = c/(c"/"100)` `(∵ "v"_e = c/100)`

= 100 = 10²