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DIA [1.3K]
3 years ago
13

Calculate the De-Broglie wavelength for electron accelerated across a potential of 10V, 100V, 1000V, and 10kV respectively. Whic

h voltage experiment exhibits a larger fringe spacing in an interference experiment is performed?
Engineering
1 answer:
Arlecino [84]3 years ago
5 0

Answer:

a) \lambda=3.87 \r A

b) \lambda=1.23 \r A

c) \lambda=0.387 \r A  

d) \lambda=0.123 \r A

Explanation:

We know that the kinetic energy can be expressed in terms of momentum (p=  mv):

K=\frac{p^{2}}{2m}

  • p is the momentum
  • m is the mass of the particle

So, the momentum will be:

p=\sqrt{2mK} (1)

We can use the energy conservation to relate K and the electric potential energy. We assume that all potential energy becomes kinetic energy. Therefore:

K=q_{e}V (2)

a) <u>If V=10 [V], K will be:</u>

K=1.6*10^{-19}*10=1.6*10^{-18}[J]

We can find the momentum using the equation (1)

p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-18}}=1.71*10^{-24}[kg*m*s^{-1}]

The De-Broglie wavelength equation is given by:

\lambda=\frac{h}{p} (3)

  • h is the Plank constant (h=6.626 x 10^{-34} [J*s])

\lambda=\frac{6.626 x 10^{-34}}{1.71*10^{-24}}=3.87*10^{-10}[m]=3.87 \r A

b) <u>If V=100 [V],</u> using the same analyze, the De-Broglie wavelength will be:

K=1.6*10^{-19}*100=1.6*10^{-17}[J]

p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-17}}=5.39*10^{-24}[kg*m*s^{-1}]

\lambda=\frac{6.626 x 10^{-34}}{5.39*10^{-24}}=1.23*10^{-10}[m]=1.23 \r A

c) <u>V=1000 [V]</u>

K=1.6*10^{-19}*100=1.6*10^{-16}[J]

p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-16}}=1.71*10^{-23}[kg*m*s^{-1}]

\lambda=\frac{6.626 x 10^{-34}}{1.71*10^{-23}}=3.87*10^{-11}[m]=0.387 \r A        

d) <u>V=10000 [V]</u>

K=1.6*10^{-19}*100=1.6*10^{-15}[J]

p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-15}}=5.40*10^{-23}[kg*m*s^{-1}]

\lambda=\frac{6.626 x 10^{-34}}{5.40*10^{-23}}=1.23*10^{-11}[m]=0.123 \r A

The fringe spacing interference is proportional to the wavelength, so in our case, the larger fringe spacing occurs when voltage is 10 V, here λ = 3.87 angstroms.

I hope it helps you!

       

 

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Answer:

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A vehicle of 1 200 kg is moving at a speed of 40 km/h on an incline of 1 in 50. The total constant rolling and wind resistance i
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