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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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A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn
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Answer:

The change in potential energy is  \Delta  PE =  -  3.8*10^{-16} \ J

Explanation:

From the question we are told that

     The  magnitude of the uniform electric field  is  E =  950 \ N/C

      The  distance traveled by the electron is  x =  2.50 \ m

Generally the force on this electron is  mathematically represented as

     F =  qE

Where F is the force and  q is the charge on the electron which is  a constant value of  q =  1.60*10^{-19} \ C

    Thus  

      F  =  950  * 1.60 **10^{-19}

      F  = 1.52 *10^{-16} \ N

Generally the work energy theorem can be mathematically represented as

          W =  \Delta  KE

Where W is the workdone on the electron by the  Electric field and  \Delta  KE  is the change in kinetic energy

Also  workdone on the electron can also  be represented as

        W =  F* x  *cos(  \theta )

Where  \theta  =  0 ^o considering that the movement of the electron is along the x-axis  

        So

             \Delta  KE  =  F  * x  cos  (0)

substituting values

         \Delta  KE  =  1.52 *10^{-16}  * 2.50   cos  (0)

          \Delta  KE   =  3.8*10^{-16} J

Now From the law of energy conservation

       \Delta PE  =  -  \Delta  KE

Where \Delta  PE is the change  in  potential energy  

Thus  

        \Delta  PE =  -  3.8*10^{-16} \ J

               

7 0
3 years ago
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