The wavelength of an electron is 10^3 nm. Wh | vedclass

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

Question

(C) According to the de-Broglie relation, the wavelength $\lambda$ is related to momentum $p$ by the formula: $\lambda = \frac{h}{p}$.Given values: $h

Vedclass · Accepted Answer

$6.625 	imes 10^{-28}$

Vedclass · Answer

(C) According to the de-Broglie relation, the wavelength $\lambda$ is related to momentum $p$ by the formula: $\lambda = \frac{h}{p}$.Given values: $h = 6.625 	imes 10^{-34} \ J \ s$ and $\lambda = 10^3 \ nm = 10^3 	imes 10^{-9} \ m = 10^{-6} \ m$.Rearranging the formula for momentum: $p = \frac{h}{\lambda}$.Substituting the values: $p = \frac{6.625 	imes 10^{-34} \ J \ s}{10^{-6} \ m} = 6.625 	imes 10^{-28} \ kg \ m \ s^{-1}$.

The wavelength of an electron is $10^3 \ nm$. What is its momentum (in $kg \ m \ s^{-1}$)? $(h = 6.625 \times 10^{-34} \ J \ s)$

Similar Questions

Which of the following represents the de Broglie equation?

If the kinetic energy of an electron of mass $9.0 \times 10^{-31} \ kg$ is $8.0 \times 10^{-25} \ J$,the wavelength of this electron in $nm$ is

The kinetic energy of an electron is $4.55 \times 10^{-25} \ J$. Find its wavelength.

Calculate the de-Broglie wavelength of an electron travelling at $1\%$ of the speed of light.

The de-Broglie wavelength of a particle with mass $1 \, g$ and velocity $100 \, m/sec$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module