A proton moving with one-tenth of the velocit | vedclass

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

Question

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.Kinetic energy $K$ is related to momentum by $K = \frac{p^

Vedclass · Accepted Answer

$4: 1$

Vedclass · Answer

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.Kinetic energy $K$ is related to momentum by $K = \frac{p^2}{2m}$.Substituting $p = \frac{h}{\lambda}$,we get $K = \frac{h^2}{2m\lambda^2}$.Since both particles have the same wavelength $\lambda$,the kinetic energy is inversely proportional to the mass: $K \propto \frac{1}{m}$.Therefore,the ratio of kinetic energy of the proton $(K_p)$ to the kinetic energy of the alpha particle $(K_\alpha)$ is $\frac{K_p}{K_\alpha} = \frac{m_\alpha}{m_p}$.Given that the mass of an alpha particle is approximately $4$ times the mass of a proton $(m_\alpha \approx 4m_p)$,we have $\frac{K_p}{K_\alpha} = \frac{4m_p}{m_p} = 4:1$.

$A$ proton moving with one-tenth of the velocity of light has a certain de Broglie wavelength of $\lambda$. An alpha particle having a certain kinetic energy has the same de Broglie wavelength $\lambda$. The ratio of the kinetic energy of the proton to that of the alpha particle is:

Similar Questions

The ratio of momentum of an electron and an alpha particle which are accelerated from rest by a potential difference of $100\,V$ is

Write the de-Broglie hypothesis and derive the equation for the de-Broglie wavelength.

$A$ proton and a photon both have the same energy of $E = 100 \,keV$. If the de Broglie wavelength of the proton is $\lambda_1$ and that of the photon is $\lambda_2$,then $\lambda_1/\lambda_2$ is proportional to:

Two particles have the same charge. If they are accelerated through the same potential difference,what will be the ratio of their de Broglie wavelengths?

$A$ 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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module