A proton and an alpha-particle are accelerate | 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 $\lambda = \frac{h}{\sqrt{2mE_k}}$,where $h$ is Planck's constant,$m$ is the mass of t

Vedclass · Accepted Answer

$2:1$

Vedclass · Answer

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE_k}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $E_k$ is the kinetic energy.Since both particles have the same kinetic energy $E_k$,we have $\lambda \propto \frac{1}{\sqrt{m}}$.Therefore,$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}}{m_p}}$.We know that the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton,i.e.,$m_{\alpha} = 4m_p$.Substituting this value,we get $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m_p}{m_p}} = \sqrt{4} = 2$.Thus,the ratio $\lambda_p : \lambda_{\alpha}$ is $2:1$.

$A$ proton and an $\alpha$-particle are accelerated from rest to the same kinetic energy. The ratio of their de Broglie wavelengths $\lambda_{p} : \lambda_{\alpha}$ is:

Similar Questions

With what potential an electron should be accelerated so that its de Broglie wavelength becomes equal to the wavelength of the first line of the Lyman series for the $He^+$ ion?

An electron microscope works on the principle of......

The de Broglie wavelength associated with a proton under the influence of an electric potential of $100 \ V$ is

Calculate the de Broglie wavelength of an electron with a kinetic energy of $200 \ eV$. [Mass of electron $= 1 \times 10^{-30} \ kg$,charge on electron $= 1.6 \times 10^{-19} \ C$,Planck's constant $(h) = 6.6 \times 10^{-34} \ Js$].

What is the ratio of the de Broglie wavelengths of a deuteron and a proton accelerated through the same potential difference?

Vedclass Test Series

Exam Paper Generator

Online Exam Module