What is the de-Broglie wavelength of the alph | vedclass

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

Question

(C) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K = qV$ is the kinetic energy.For an $\alpha

Vedclass · Accepted Answer

$\frac{0.101}{\sqrt{V}} \ \mathring{A}$

Vedclass · Answer

(C) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K = qV$ is the kinetic energy.For an $\alpha$-particle,the charge $q = 2e = 2 	imes 1.6 	imes 10^{-19} \ C$ and mass $m = 4m_p = 4 	imes 1.67 	imes 10^{-27} \ kg$.Substituting these values:$\lambda = \frac{h}{\sqrt{2(4m_p)(2e)V}} = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 4 	imes 1.67 	imes 10^{-27} 	imes 2 	imes 1.6 	imes 10^{-19} 	imes V}}$$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{42.752 	imes 10^{-46} 	imes V}} = \frac{6.63 	imes 10^{-34}}{6.538 	imes 10^{-23} \sqrt{V}} \ m$$\lambda \approx \frac{1.014 	imes 10^{-11}}{\sqrt{V}} \ m = \frac{0.1014 	imes 10^{-10}}{\sqrt{V}} \ m = \frac{0.101}{\sqrt{V}} \ \mathring{A}$.

What is the de-Broglie wavelength of the $\alpha$-particle accelerated through a potential difference $V$?

Similar Questions

The de-Broglie wavelength of a particle moving with a velocity $2.25 \times 10^8\, m/s$ is equal to the wavelength of a photon. The ratio of the kinetic energy of the particle to the energy of the photon is (velocity of light is $3 \times 10^8\, m/s$). (in $/8$)

Let $E_e$ and $E_p$ represent the kinetic energy of an electron and a photon,respectively. If the de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron,then find the ratio $E_p / E_e$. (Given: speed of electron $v = c / 100$,where $c$ is the velocity of light).

$A$ subatomic particle of mass $10^{-30} \ kg$ is moving with a velocity $2.21 \times 10^6 \ m/s$. Under the matter wave consideration,the particle will behave closely like . . . . . . . $(h = 6.63 \times 10^{-34} \ J \cdot s)$

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 idea of matter waves was given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module