The de Broglie wavelength of a proton and alp | vedclass

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

Question

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the

Vedclass · Accepted Answer

$4:1$

Vedclass · Answer

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the velocity.Given that the de Broglie wavelengths are equal,$\lambda_p = \lambda_{\alpha}$.Therefore,$\frac{h}{m_p v_p} = \frac{h}{m_{\alpha} v_{\alpha}}$.This implies $m_p v_p = m_{\alpha} v_{\alpha}$.We know that the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton,so $m_{\alpha} = 4m_p$.Substituting this into the equation: $m_p v_p = (4m_p) v_{\alpha}$.Dividing both sides by $m_p$,we get $v_p = 4v_{\alpha}$.Thus,the ratio of their velocities is $\frac{v_p}{v_{\alpha}} = 4:1$.

The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities 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).

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 $\frac{K}{4}$?

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelength of the electron to the wavelength of the photon is ($c =$ velocity of light).

Two particles move at right angles to each other. Their de Broglie wavelengths are $\lambda_1$ and $\lambda_2$ respectively. The particles undergo a perfectly inelastic collision. The de Broglie wavelength $\lambda$ of the final particle is given by

The speed of electrons in a scanning electron microscope is $1 \times 10^{7} \, m/s$. If protons having the same speed are used instead of electrons,then the resolving power of the scanning proton microscope will be changed by a factor of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module