Question

If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πm × 0.05 nm, is there any problem in defining this value.​

Answer 1

Answer :

• Uncertainty in momentum = 2.638 × 10⁻²³ kg ms⁻¹.

Step-by-step explanation :

Given that,

• Uncertainty in the position of the electron, Δx = 0.002 nm = 2 × 10⁻³ nm = 2 × 10⁻¹² m.

Also,

• Planck's constant, h = 6.626 × 10⁻³⁴ kgm²s⁻¹.

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Using uncertainty principle, we get,

$\implies\rm{{\Delta}x.{\Delta}p=\dfrac{h}{4\pi}}$

$\implies\rm{{\Delta}p=\dfrac{h}{4\pi{\Delta}x}}$

Putting the given values, we get,

$\implies\rm{\Delta\:p=\dfrac{6.626\times{}10^{-34}\:kgm^2s^{-1}}{4\times{}3.14\times{}(2\times{}10^{-12}\:m)}}$

$\implies\rm{\Delta\:p=}\:\bold{2.638\times{}10^{-23}\:kg\:ms^{-1}}$

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Now, actual momentum (as per given in the question) =

$\implies\rm{\dfrac{h}{4\pi\times{}0.05\:nm}=\dfrac{h}{4\pi\times{}5\times{}10^{-11}\:m}}$

Putting the values, we get,

$\implies\rm{\dfrac{6.626\times{}10^{-34}\:kgm^2s^{-1}}{4\times{}3.14\times{}10^{-11}\:m}}$

$\implies\rm{1.055\times{}10^{-24}\:kgms^{-1}}$

Hence, it cannot be defined as the actual magnitude of momentum is smaller than the uncertainty.

Answer 2

Answer:

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