The de Broglie wavelength is given as $1 \, \mathring{A}$ and the value of $h$ is $6.6252 \times 10^{-27} \, \text{erg} \cdot \text{s}$. What will be the momentum of the particle (in $\text{g} \cdot \text{cm/s}$)?
- A$6.6252 \times 10^{-19}$
- B$6.6252 \times 10^{-21}$
- C$6.6252 \times 10^{-24}$
- D$6.6252 \times 10^{-27}$
Explore More
Similar Questions
The wavelength of a microscopic particle of mass $9.1 \times 10^{-31} \ kg$ is $182 \ nm$. Its kinetic energy in $J$ is: $(h = 6.625 \times 10^{-34} \ J \ s)$
If the velocity of a golf ball with a mass of $200 \ g$ is $5 \ m/hr$,then the wavelength associated with it will be .........
Difficult
View SolutionIf the velocity of a hydrogen molecule is $5 \times 10^4 \, cm \, sec^{-1}$,then its de-Broglie wavelength is ............. $\mathring{A}$.
Difficult
View SolutionThe velocity associated with a proton moving in a potential difference of $1000 \, V$ is $4.37 \times 10^{5} \, ms^{-1}$. If a hockey ball of mass $0.1 \, kg$ is moving with this velocity,calculate the wavelength associated with this velocity.
Medium
View SolutionThe basis of the quantum mechanical model of an atom is:
Vedclass 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