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 de-Broglie's wavelength of an electron in the $4^{th}$ orbit is . . . . . . . . $\pi a_0$. ($a_0 =$ Bohr's radius)

If $\lambda _L, \lambda _M$ and $\lambda _N$ are the wavelengths of an electron in $L, M, N$ energy levels of $H^{-}$ atom respectively,what is their decreasing order?

An electron has kinetic energy $2.8 \times 10^{-23} \ J$. The de-Broglie wavelength will be nearly $(m_e = 9.1 \times 10^{-31} \ kg)$.

The energy of separation of an electron in a Hydrogen-like atom in an excited state is $3.4 \ eV$. The de-Broglie wavelength (in $\mathring{A}$) associated with the electron is (Given: radius of the first orbit of $H$ atom is $0.53 \ \mathring{A}$)

The frequency of the de-Broglie wave of an electron in Bohr's first orbit of a hydrogen atom is . . . . . . . . . . $\times 10^{13} \ Hz$ (nearest integer).
[Given: $R_H$ (Rydberg constant) $= 2.18 \times 10^{-18} \ J$,$h$ (Planck's constant) $= 6.6 \times 10^{-34} \ J \cdot s$]

Vedclass Test Series

Exam Paper Generator

Online Exam Module