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VLD [36.1K]
3 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]3 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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Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to
sergij07 [2.7K]

Answer:

KE=1.2036\times 10^{-12}\ J

Explanation:

Given:

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<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

q.v.B=m.\frac{v^2}{r}

m.v=q.B.r

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v=\frac{q.B.r}{m}

v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}

v=1.2048\times 10^{7}\ m.s^{-1}

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

<u>We firstly find the relativistic mass as:</u>

m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m

m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }

m'=6.6533\times10^{-27}\ kg

now kinetic energy:

KE=m'.c-m.c

KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2

KE=1.2036\times 10^{-12}\ J

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