Question

Calculate the De-Broglie wavelength for electron accelerated across a potential of 10V, 100V, 1000V, and 10kV respectively. Which voltage experiment exhibits a larger fringe spacing in an interference experiment is performed?

Answer 1

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) If V=10 [V], K will be:

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

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) If V=100 [V], 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) V=1000 [V]

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

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

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

d) V=10000 [V]

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

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

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

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!