An electron,an alpha-particle,and a proton ha | vedclass

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

Question

(C) The de Broglie wavelength $\lambda$ of a particle is given by $\lambda = h / p$.Since kinetic energy $K = p^2 / (2m)$,we can write momentum $p = \

Vedclass · Accepted Answer

$\alpha$-particle

Vedclass · Answer

(C) The de Broglie wavelength $\lambda$ of a particle is given by $\lambda = h / p$.Since kinetic energy $K = p^2 / (2m)$,we can write momentum $p = \sqrt{2mK}$.Substituting this into the wavelength formula,we get $\lambda = h / \sqrt{2mK}$.For a constant kinetic energy $K$,the wavelength is inversely proportional to the square root of the mass: $\lambda \propto 1 / \sqrt{m}$.Comparing the masses: $m_{\alpha} > m_p > m_e$.Since the $\alpha$-particle has the largest mass among the three,it will have the smallest de Broglie wavelength.

An electron,an $\alpha$-particle,and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?

Similar Questions

The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^8 \ m/s$ is equal to that of a photon. What is the ratio of the kinetic energy of the electron to that of the photon? (Given: $c = 3 \times 10^8 \ m/s$)

The de-Broglie wavelength of a body of mass $1 \ kg$ moving with a velocity of $2000 \ m/s$ is:

The de Broglie wavelength of a molecule in a gas at room temperature $(300 \ K)$ is $\lambda_1$. If the temperature of the gas is increased to $600 \ K$,then the de Broglie wavelength of the same gas molecule becomes $..............$.

The energy that should be added to an electron to reduce its de Broglie wavelength from $1 \, nm$ to $0.5 \, nm$ is

An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are moving with the same speed. The ratio of their de-Broglie's wavelengths $\lambda_{e} / \lambda_{p}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module