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  • An electron with a momentum of <img alt="2.5 \times 10^{-24} \mathrm{kgm} / \mathrm{s}" src="https://entrancecorner.oncodecogs.com/gif.latex?2.5%20%5Ctimes%2010%5E%7B-24%7D%20%5Cmathrm%7Bkgm%7

An electron with a momentum of 2.5 \times 10^{-24} \mathrm{kgm} / \mathrm{s} has a de Broglie wavelength of:

 

Option: 1

2.65\times 10^{-10}\; m


Option: 2

4.37\times 10^{-12}\; m


Option: 3

1.06\times 10^{-11}\; m


Option: 4

3.54\times 10^{-12}\; m


Answers (1)

best_answer

The de Broglie wavelength of a particle is given by \lambda=\frac{h}{p} , 

where h is Planck's constant and p is the momentum of the particle. 

Substituting the given value of momentum, we get

\lambda=\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{2.5 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s}} \approx 2.65 \times 10^{-10} \mathrm{~m}

 

 

 

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