Answer :
Answer:
a) [tex]\lambda=3.87 \r A[/tex]
b) [tex]\lambda=1.23 \r A[/tex]
c) [tex]\lambda=0.387 \r A[/tex]
d) [tex]\lambda=0.123 \r A[/tex]
Explanation:
We know that the kinetic energy can be expressed in terms of momentum (p= mv):
[tex]K=\frac{p^{2}}{2m}[/tex]
- p is the momentum
- m is the mass of the particle
So, the momentum will be:
[tex]p=\sqrt{2mK}[/tex] (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:
[tex]K=q_{e}V[/tex] (2)
a) If V=10 [V], K will be:
[tex]K=1.6*10^{-19}*10=1.6*10^{-18}[J][/tex]
We can find the momentum using the equation (1)
[tex]p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-18}}=1.71*10^{-24}[kg*m*s^{-1}][/tex]
The De-Broglie wavelength equation is given by:
[tex]\lambda=\frac{h}{p}[/tex] (3)
- h is the Plank constant ([tex]h=6.626 x 10^{-34} [J*s][/tex])
[tex]\lambda=\frac{6.626 x 10^{-34}}{1.71*10^{-24}}=3.87*10^{-10}[m]=3.87 \r A[/tex]
b) If V=100 [V], using the same analyze, the De-Broglie wavelength will be:
[tex]K=1.6*10^{-19}*100=1.6*10^{-17}[J][/tex]
[tex]p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-17}}=5.39*10^{-24}[kg*m*s^{-1}][/tex]
[tex]\lambda=\frac{6.626 x 10^{-34}}{5.39*10^{-24}}=1.23*10^{-10}[m]=1.23 \r A[/tex]
c) V=1000 [V]
[tex]K=1.6*10^{-19}*100=1.6*10^{-16}[J][/tex]
[tex]p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-16}}=1.71*10^{-23}[kg*m*s^{-1}][/tex]
[tex]\lambda=\frac{6.626 x 10^{-34}}{1.71*10^{-23}}=3.87*10^{-11}[m]=0.387 \r A[/tex]
d) V=10000 [V]
[tex]K=1.6*10^{-19}*100=1.6*10^{-15}[J][/tex]
[tex]p=\sqrt{2*9.1 x 10^{-31}*1.6*10^{-15}}=5.40*10^{-23}[kg*m*s^{-1}][/tex]
[tex]\lambda=\frac{6.626 x 10^{-34}}{5.40*10^{-23}}=1.23*10^{-11}[m]=0.123 \r A[/tex]
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!