The de-Broglie wavelength of a particle of ma | 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 $\lambda = \frac{h}{mv}$.Given:Mass $m = 6.63 \ g = 6.63 	imes 10^{-3} \ kg$Velocity

Vedclass · Accepted Answer

$10^{-33} \ m$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.Given:Mass $m = 6.63 \ g = 6.63 	imes 10^{-3} \ kg$Velocity $v = 100 \ ms^{-1}$Planck's constant $h = 6.63 	imes 10^{-34} \ J \cdot s$Substituting the values:$\lambda = \frac{6.63 	imes 10^{-34} \ J \cdot s}{(6.63 	imes 10^{-3} \ kg) 	imes (100 \ ms^{-1})}$$\lambda = \frac{6.63 	imes 10^{-34}}{6.63 	imes 10^{-1}} \ m$$\lambda = 10^{-33} \ m$

The de-Broglie wavelength of a particle of mass $6.63 \ g$ moving with a velocity of $100 \ ms^{-1}$ is

Similar Questions

What is the characteristic of an electron microscope? On which principle is it based?

Which one of the following explains light both as a stream of particles and as wave motion?

Which of the following relations is correct, if the wavelength $(\lambda)$ is equal to the distance travelled by the electron in one second? $h$ is the Planck's constant and $m$ is the mass of electron.

Dual nature of particles was proposed by

Calculate the de Broglie wavelength of a $CO_2$ molecule moving with a velocity of $440 \, m/s$. (Molar mass of $CO_2 = 44 \, g/mol$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module