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2 years ago

What is the energy of a single photon of visible light with a wavelength of 500.0 nm? 1 nm=10^−9 m

Physics

1 answer:

Dmitrij [34]2 years ago

8 0

Answer:

3.97×10^−19 J

From the planks equation

E=hv

V= c/ wave length

V= 3×10^8/500×10^-9

=6×10^14

E= hv

6.63×10^-34×6×10^14

= 3.97×10^−19 J

Explanation:

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An object is moving along a straight line, and the uncertainty in its position is 1.90 m.

just olya [345]

Answer:

$2.78\times 10^{-35}\ \text{kg m/s}$

$6.178\times 10^{-34}\ \text{m/s}$

$0.31\times 10^{-4}\ \text{m/s}$

Explanation:

$\Delta x$ = Uncertainty in position = 1.9 m

$\Delta p$ = Uncertainty in momentum

h = Planck's constant = $6.626\times 10^{-34}\ \text{Js}$

m = Mass of object

From Heisenberg's uncertainty principle we know

$\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}$

The minimum uncertainty in the momentum of the object is $2.78\times 10^{-35}\ \text{kg m/s}$

Golf ball minimum uncertainty in the momentum of the object

$m=0.045\ \text{kg}$

Uncertainty in velocity is given by

$\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}$

The minimum uncertainty in the object's velocity is $6.178\times 10^{-34}\ \text{m/s}$

Electron

$m=9.11\times 10^{-31}\ \text{kg}$

$\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}$