The de Broglie wavelength for a deuteron can | vedclass

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

Question

(C) The de Broglie wavelength $\lambda$ associated with a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given

Vedclass · Accepted Answer

$\lambda _d = \frac{0.202}{\sqrt{V}} \ \mathring{A}$

Vedclass · Answer

(C) The de Broglie wavelength $\lambda$ associated with a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For a deuteron, the mass $m_d \approx 2m_p$ (where $m_p$ is the mass of a proton) and the charge $q_d = e$.Substituting the values: $m_d = 2 	imes 1.67 	imes 10^{-27} \ 	ext{kg} = 3.34 	imes 10^{-27} \ 	ext{kg}$ and $q_d = 1.6 	imes 10^{-19} \ 	ext{C}$.Using the formula $\lambda = \frac{h}{\sqrt{2m_d e V}}$, we get $\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 3.34 	imes 10^{-27} 	imes 1.6 	imes 10^{-19} 	imes V}}$.Calculating the constant: $\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{10.688 	imes 10^{-46} 	imes V}} = \frac{6.63 	imes 10^{-34}}{3.27 	imes 10^{-23} \sqrt{V}} \approx \frac{2.02 	imes 10^{-11}}{\sqrt{V}} \ 	ext{m}$.Converting to $\mathring{A}$ $(1 \ \mathring{A} = 10^{-10} \ 	ext{m})$: $\lambda = \frac{0.202}{\sqrt{V}} \ \mathring{A}$.

The de Broglie wavelength for a deuteron can be given by:

Similar Questions

An electron with mass $m$ and an initial velocity $(t=0)$ $\vec{v} = v_0 \hat{i}$ $(v_0 > 0)$ enters a magnetic field $\vec{B} = B_0 \hat{j}$. If the initial de Broglie wavelength at $t=0$ is $\lambda_0$,then its value after time $t$ would be:

The accelerating voltage of an electron gun is $50,000 \ V$. What will be the de Broglie wavelength of the electron in $\mathring{A}$?

The voltage applied to an electron microscope to produce electrons of wavelength $0.50 \text{ Å}$ is (in $\text{ V}$)

The wavelength $\lambda$ of a photon and the de-Broglie wavelength of an electron have the same value. The ratio of the kinetic energy of the electron to the energy of a photon is ($m=$ mass of electron,$c=$ velocity of light,$h=$ Planck's constant).

Compute the typical de Broglie wavelength of an electron in a metal at $27\,^{\circ} C$ and compare it with the mean separation between two electrons in a metal which is given to be about $2 \times 10^{-10} \; m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module