If an electron is accelerated by a potential | vedclass

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

Question

(A) $1$. Kinetic Energy $(KE)$: $KE = e 	imes V = (1.602 	imes 10^{-19} \ C) 	imes (1.0 	imes 10^4 \ V) = 1.602 	imes 10^{-15} \ J$. $2$. Frequen

Vedclass · Accepted Answer

$KE = 1.602 	imes 10^{-15} \ J, 
u = 2.418 	imes 10^{18} \ Hz, \lambda = 1.227 	imes 10^{-11} \ m$

Vedclass · Answer

(A) $1$. Kinetic Energy $(KE)$: $KE = e 	imes V = (1.602 	imes 10^{-19} \ C) 	imes (1.0 	imes 10^4 \ V) = 1.602 	imes 10^{-15} \ J$. $2$. Frequency $(
u)$: Using $KE = h
u$,$
u = \frac{KE}{h} = \frac{1.602 	imes 10^{-15} \ J}{6.626 	imes 10^{-34} \ J \cdot s} = 2.418 	imes 10^{18} \ Hz$. $3$. Wavelength $(\lambda)$: Using de Broglie equation $\lambda = \frac{h}{\sqrt{2m(KE)}}$,$\lambda = \frac{6.626 	imes 10^{-34}}{\sqrt{2 	imes 9.11 	imes 10^{-31} 	imes 1.602 	imes 10^{-15}}} = 1.227 	imes 10^{-11} \ m$.

If an electron is accelerated by a potential difference of $1.0 \times 10^4 \ V$,calculate its kinetic energy,frequency,and wavelength.

Similar Questions

The velocity associated with a proton moving in a potential difference of $1000 \, V$ is $4.37 \times 10^{5} \, ms^{-1}$. If a hockey ball of mass $0.1 \, kg$ is moving with this velocity,calculate the wavelength associated with this velocity.

The velocity of an electron is $1\%$ of the velocity of light. Calculate the de-Broglie wavelength of the electron.

Dual nature of particles was proposed by

Find the number of waves formed by an electron in the $n = 3$ orbit of a Bohr atom.

Vedclass Test Series

Exam Paper Generator

Online Exam Module