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rusak2 [61]
3 years ago
12

For a ∆x of 0.1mm, what is ∆px, the uncertainty in the transverse momentum of a photon passing through a slit (where uncertainty

in x comes from slit width, ∆x=slit width?
Physics
1 answer:
Colt1911 [192]3 years ago
6 0

Answer:

0.53\times 10^{-30}kgms^{-1}

Explanation:

Uncertainty principle say that the position and momentum can not be measured simultaneously except one relation which is described below,

\Delta x\Delta p=\frac{h}{4\pi }

Given that the uncertainty in x is 0.1 mm.

Therefore,

\Delta p=\frac{6.626\times 10^{-34} }{4\times 3.14\times 1\times 10^{-4}m }\\\Delta p=0.53\times 10^{-30}kgms^{-1}

Therefore, uncertainty in the transverse momentum of photon is 0.53\times 10^{-30}kgms^{-1}

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0.37sec

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An oscillating pendulum, or anything else in nature that involves "simple harmonic" (sinusoidal) motion, spends 1/4 of its period going from zero speed to maximum speed, and another 1/4 going from maximum speed to zero speed again, etc. After four quarter-periods it is back where it started.

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A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?
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Answer:

U= 0.112 x10^{-6} J

u= 0.708 \frac{J}{m^{3} }

Explanation:

diameter= 2 cm

r= 1 cm * \frac{1m}{100cm} = 0.01 m

distance= 0.5 mm

d= 0.5 mm * \frac{1m}{1000m}= 0.5 x10^{-3}

Area A= \pi *r^{2}= 0.314x10^{-3}  m^{2}

Volume v= A*d = 0.314 x10^{-3}m^{2}*0.5x10^{-3} m =0.157x10^{-6}  m^{3}

v= 0.157 x10^{-6} m^{3}

Constant vacuum permittivity

E_{o}= 8.85x10^{-12}

a).

C= \frac{A*E_{o} }{d} = \frac{0.314x10^{-3} *8.85x10^{-12}  }{0.5x10^{-3} } \\C= 5.56 x10^{-12}F

U= \frac{1}{2} *C *V^{2}\\ U=\frac{1}{2} * 5.56x10^{-12}*(200)^{2}  \\U=0.112 x^{-6} J\\

b).

u=\frac{U}{v}

u=\frac{0.112 x10^{-6}J}{0.157x10^{-6}m^{3}  } \\u=0.708 \frac{J}{m^{3} }

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