If the de Broglie wavelength of an electron i | vedclass

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

Question

(D) The de Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{mv}$.The wavelength of a photon is given by $\lambda_p = \frac{c}{
u}$

Vedclass · Accepted Answer

$1.45 	imes 10^6 \, m/s$

Vedclass · Answer

(D) The de Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{mv}$.The wavelength of a photon is given by $\lambda_p = \frac{c}{
u}$,where $
u$ is the frequency.According to the problem,$\lambda_e = 10^{-3} 	imes \lambda_p$.Substituting the expressions: $\frac{h}{mv} = 10^{-3} 	imes \frac{c}{
u}$.Rearranging for the speed of the electron $(v)$: $v = \frac{h 
u}{m c 	imes 10^{-3}}$.Substituting the given values: $v = \frac{6.63 	imes 10^{-34} 	imes 6 	imes 10^{14}}{9.1 	imes 10^{-31} 	imes 3 	imes 10^8 	imes 10^{-3}}$.$v = \frac{39.78 	imes 10^{-20}}{27.3 	imes 10^{-26}} = 1.457 	imes 10^6 \, m/s \approx 1.45 	imes 10^6 \, m/s$.

Similar Questions

What is the de Broglie wavelength associated with
$(a)$ an electron moving with a speed of $5.4 \times 10^{6} \; m/s$,and
$(b)$ a ball of mass $150 \; g$ travelling at $30.0 \; m/s$?

The graph shows the variation of de-Broglie wavelength $\lambda$ versus $\frac{1}{\sqrt{V}}$,where $V$ is the accelerating potential for four particles carrying the same charge but having masses $m_1, m_2, m_3, m_4$. Which one represents the particle with the smallest mass?

The work done to accelerate an electron from rest so that it can have a de Broglie wavelength of $6600 \text{ Å}$ is nearly (Planck's constant $= 6.6 \times 10^{-34} \text{ J s}$ and mass of electron $= 9 \times 10^{-31} \text{ kg}$)

An electron is accelerated from rest through a potential difference of $V$ volt. If the de Broglie wavelength of the electron is $1.227 \times 10^{-2} \, nm$,the potential difference is $......V$.

An electron of mass $m$ when accelerated through a potential difference $V$ has a de-Broglie wavelength $\lambda$. The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module