The wavelength of an electron is $10^3 \ nm$. What is its momentum (in $kg \ m \ s^{-1}$)? $(h = 6.625 \times 10^{-34} \ J \ s)$
- A$6.625 \times 10^{-31}$
- B$6.625 \times 10^{-37}$
- C$6.625 \times 10^{-28}$
- D$6.625 \times 10^{-34}$
Explore More
Similar Questions
The mass of a proton is $1.67 \times 10^{-27} \ kg$. If it is moving with a velocity of $10^3 \ m/s$,calculate its wavelength in nanometers.
Medium
View SolutionIf the de Broglie wavelength of an electron is $728.14 \ nm$,its kinetic energy in $J$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)
If an electron,hydrogen nucleus,helium nucleus,and neon nucleus are all moving with the same velocity,then the wavelengths associated with these particles are in the order:
Medium
View SolutionIf the kinetic energy of a particle is increased to $4$ times its initial value,how many times will the associated de Broglie wavelength become?
Difficult
View SolutionThe kinetic energy (in $J$) of a particle of mass $4.5 \times 10^{-31} \ kg$ having a wavelength of $1000 \ nm$ is : $(h = 6.62 \times 10^{-34} \ J \ s)$
DifficultAP EAMCET 2017
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo