The de Broglie wavelength associated with a b | vedclass

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

Question

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $E = \frac{p^2}

Vedclass · Accepted Answer

$6.626 	imes 10^{-34} \, m$

Vedclass · Answer

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.Substituting this into the de Broglie equation: $\lambda = \frac{h}{\sqrt{2mE}}$.Given $m = 1 \, kg$,$E = 0.5 \, J$,and $h = 6.626 	imes 10^{-34} \, J \cdot s$:$\lambda = \frac{6.626 	imes 10^{-34}}{\sqrt{2 	imes 1 	imes 0.5}} = \frac{6.626 	imes 10^{-34}}{\sqrt{1}} = 6.626 	imes 10^{-34} \, m$.

The de Broglie wavelength associated with a ball of mass $1 \, kg$ having kinetic energy $0.5 \, J$ is

Similar Questions

If the wavelength and the distance traveled by an electron in $1 \ s$ are the same,calculate its velocity.

If the de Broglie wavelength of an electron is $728.14 \ nm$,its kinetic energy in $J$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)

$A$ proton accelerated through a potential $V$ has de-Broglie wavelength $\lambda$. Then the de-Broglie wavelength of an $\alpha$-particle,when accelerated through the same potential $V$ is :

Two particles $A$ and $B$ are in motion. If the wavelength associated with particle $A$ is $5 \times 10^{-8} \ m$,calculate the wavelength (in $\mathring{A}$) associated with particle $B$ if its momentum is half of $A$.

Calculate the momentum of a particle which has a wavelength of $1 \ \mathring{A}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module