If the de-Broglie wavelengths for a proton an | vedclass

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

Question

(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Given that the de-Broglie wavelengths for a proton $(p)$ and an $\alpha$-particle $

Vedclass · Accepted Answer

$4 : 1$

Vedclass · Answer

(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Given that the de-Broglie wavelengths for a proton $(p)$ and an $\alpha$-particle $(\alpha)$ are equal, we have $\lambda_p = \lambda_{\alpha}$.Therefore, $\frac{h}{m_p v_p} = \frac{h}{m_{\alpha} v_{\alpha}}$, which simplifies to $m_p v_p = m_{\alpha} v_{\alpha}$.The ratio of their velocities is $\frac{v_p}{v_{\alpha}} = \frac{m_{\alpha}}{m_p}$.We know that the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton, so $m_{\alpha} = 4 m_p$.Substituting this, we get $\frac{v_p}{v_{\alpha}} = \frac{4 m_p}{m_p} = \frac{4}{1}$.Thus, the ratio of their velocities is $4 : 1$.

If the de-Broglie wavelengths for a proton and for an $\alpha$-particle are equal, then the ratio of their velocities will be

Similar Questions

$A$ particle is moving $5$ times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is $1.878 \times 10^{-4}$. The mass of the particle is close to

The wavelength of the matter wave is independent of

If the kinetic energy of an electron is increased four times,the wavelength of the de-Broglie wave associated with it would become

The idea of matter waves was given by

The graph which shows the variation of the de Broglie wavelength $(\lambda)$ of a particle and its associated momentum $(p)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module