A particle of mass 1, mg has the same wavelen | vedclass

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

Question

(C) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Since the particle and the electron have the same wavelength,their momenta must be

Vedclass · Accepted Answer

$2.7 	imes 10^{-18}\, m\,s^{-1}$

Vedclass · Answer

(C) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Since the particle and the electron have the same wavelength,their momenta must be equal.$m_p v_p = m_e v_e$Here,$m_p = 1\, mg = 10^{-6}\, kg$,$m_e = 9.1 	imes 10^{-31}\, kg$,and $v_e = 3 	imes 10^6\, m\,s^{-1}$.Substituting the values:$v_p = \frac{m_e v_e}{m_p} = \frac{9.1 	imes 10^{-31} 	imes 3 	imes 10^6}{10^{-6}}$$v_p = 27.3 	imes 10^{-25} 	imes 10^6 = 27.3 	imes 10^{-19} = 2.73 	imes 10^{-18}\, m\,s^{-1}$.Rounding to the given options,the velocity is $2.7 	imes 10^{-18}\, m\,s^{-1}$.

$A$ particle of mass $1\, mg$ has the same wavelength as an electron moving with a velocity of $3 \times 10^6\, m\,s^{-1}$. The velocity of the particle is (Mass of electron $= 9.1 \times 10^{-31}\, kg$)

Similar Questions

If the de Broglie wavelengths of an electron and a proton are equal,then what will be the kinetic energy of the electron?

The de-Broglie wavelength of a particle of kinetic energy $K$ is $\lambda$. What will be the wavelength of the particle if its kinetic energy becomes $\frac{K}{4}$?

Proton $(P)$ and electron $(e)$ will have the same de Broglie wavelength when the ratio of their momentum is (assume,$m_{p} = 1849 \, m_{e}$)

$A$ particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head-on and elastic. The ratio of the de-Broglie wavelengths $\lambda_A$ and $\lambda_B$ after the collision is

The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is $E$. If $\lambda_1$ and $\lambda_2$ are the de Broglie wavelengths of the proton and the photon respectively,then find the ratio $\lambda_1 / \lambda_2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module