An alpha particle and a carbon 12 atom have t | vedclass

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

Question

(B) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $K = \frac{p^2}{2m}$,we hav

Vedclass · Accepted Answer

$\sqrt{3} : 1$

Vedclass · Answer

(B) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $K = \frac{p^2}{2m}$,we have $p = \sqrt{2mK}$.Substituting this into the wavelength formula: $\lambda = \frac{h}{\sqrt{2mK}}$.For the same kinetic energy $K$,$\lambda \propto \frac{1}{\sqrt{m}}$.Therefore,$\frac{\lambda_{\alpha}}{\lambda_{C12}} = \sqrt{\frac{m_{C12}}{m_{\alpha}}}$.The mass of an $\alpha$ particle is approximately $4 	ext{ amu}$ and the mass of a carbon $12$ atom is $12 	ext{ amu}$.$\frac{\lambda_{\alpha}}{\lambda_{C12}} = \sqrt{\frac{12}{4}} = \sqrt{3} = \sqrt{3} : 1$.

An $\alpha$ particle and a carbon $12$ atom have the same kinetic energy $K$. The ratio of their de-Broglie wavelengths $(\lambda_{\alpha} : \lambda_{C12})$ is

Similar Questions

The de-Broglie wavelength of an electron moving with a velocity $v = c / 2$ ($c$ is the velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of the electron and the photon is:

$A$ photon and an electron (mass $m$) have the same energy $E$. The ratio $\left(\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}}\right)$ of their de Broglie wavelengths is: ($c$ is the speed of light)

State the de-Broglie hypothesis.

Electrons are accelerated through a potential difference of $16 \ kV$. If the potential difference is increased to $64 \ kV$,then the de-Broglie wavelength associated with the electron will

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module