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VLD [36.1K]
2 years ago
6

The speed of a certain proton is 350 km/s. If the uncertainty in its momentum is 0.100%, what is the necessary uncertainty in it

s location?
Physics
1 answer:
Evgesh-ka [11]2 years ago
5 0

Answer:

\Delta x = 1.807 \times 10^{-10}m

Explanation:

mass of proton, m = 1.67 x 10^-27 kg

speed of proton, v = 350 km/s = 350,000 m/s

Momentum of proton, p = mass x speed

p =  1.67 x 10^-27 x 350000 = 5.845 x 10^-22 kg m /s

uncertainty in momentum, Δp = 0.1 % of p

Δp = \frac{0.1\times 5.845 \times 10^{-22}}{100}=5.845 \times 10^{-25}

According to the principle

\Delta x\times \Delta p \geq  \frac{h}{2\pi }

where, Δx be the uncertainty in position

\Delta x\times 5.845 \times 10^{-25}=  \frac{6.634 \times 10^{-34}}{2\times 3.14}

\Delta x = 1.807 \times 10^{-10}m

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