The de Broglie wavelength of an electron movi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that the de Broglie wavelength of the electron $\lambda_e$ is equal to the wavelength of the photon $\lambda_p$,so $\lambda_e = \lambda_p =

Vedclass · Accepted Answer

$1/4$

Vedclass · Answer

(A) Given that the de Broglie wavelength of the electron $\lambda_e$ is equal to the wavelength of the photon $\lambda_p$,so $\lambda_e = \lambda_p = \lambda$.The kinetic energy of the electron is $K_e = \frac{1}{2} m_e v_e^2$.The energy of the photon is $E_p = \frac{hc}{\lambda_p}$.From the de Broglie relation,$\lambda_e = \frac{h}{m_e v_e}$,which implies $m_e = \frac{h}{\lambda_e v_e}$.Substituting $m_e$ into the kinetic energy expression: $K_e = \frac{1}{2} \left( \frac{h}{\lambda_e v_e} ight) v_e^2 = \frac{h v_e}{2 \lambda_e}$.Now,the ratio of the kinetic energy of the electron to the energy of the photon is:$\frac{K_e}{E_p} = \frac{h v_e / 2 \lambda_e}{hc / \lambda_p} = \frac{v_e}{2c}$.Given $v_e = 1.5 	imes 10^8 \, m/s$ and $c = 3 	imes 10^8 \, m/s$:$\frac{K_e}{E_p} = \frac{1.5 	imes 10^8}{2 	imes 3 	imes 10^8} = \frac{1.5}{6} = \frac{1}{4}$.

The de Broglie wavelength of an electron moving with a velocity $1.5 \times 10^8 \, m/s$ is equal to that of a photon. The ratio of the kinetic energy of the electron to the energy of the photon is

Similar Questions

How much energy is imparted to an electron so that its de-Broglie wavelength reduces from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$? (Let $E$ be the initial energy of the electron).

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is ($c$ = velocity of light in air).

If the kinetic energy of a particle is increased to $16$ times, the percentage change in the de Broglie wavelength of a particle is (in $\%$)

If the velocity of a moving particle is reduced to half, what will be the percentage change in its de Broglie wavelength?

What is the de-Broglie wavelength associated with an electron moving with a speed of $6.4 \times 10^{6} \ m/s$ (in $nm$)? $(m_{e} = 9.11 \times 10^{-31} \ kg, h = 6.63 \times 10^{-34} \ J \cdot s)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module