Respuesta :
Answer:
The width L of the barrier is b. 0.116nm
Explanation:
This is a case of tunnel effect when the probability is less than one.
First of all you find all the information the problem gives you:
Electron's total energy E=5eV
Height of potential barrier U=20eV
Crossing Probability T=0.03
Width of the barrier L=unknown
Then you need to find the equations to use in cases in which the probability is less than 0, so you have:
[tex]T=Ge^{-2kL}[/tex]
[tex]G=16\frac{E}{U}(1-\frac{E}{U})[/tex]
[tex]k=\frac{\sqrt{2m(U-E)}}{h}[/tex] where m is the mass of the electron, [tex]m=9.11*10^{-31}Kg[/tex], and h is the Planck constant, [tex]h=1.054*10^{-34}J.s[/tex]
First we find G, we have:
[tex]G=16(\frac{5.0eV}{20eV})(1-\frac{5.0eV}{20eV})[/tex]
[tex]G=4(1-0.25)=3[/tex]
Now, to solving k, we need to find the difference (U-E) :
U - E = 20eV - 5eV = 15eV
[tex]15eV*\frac{1.60218*10^{-19}J}{1eV} =2.403*10^{-18}J[/tex]
Then, we can find k:
[tex]k=\frac{\sqrt{2m(U-E)}}{h}[/tex]
[tex]k=\frac{\sqrt{2(9.11*10^{31}kg)(2.403*10^{-18}J)}}{1.054*10^{-34}\frac{J}{s}}[/tex]
[tex]k=1.98*10^{10}[/tex]
Finally, we solve for L the probability equation T:
[tex]T=Ge^{-2kL}[/tex]
[tex]\frac{T}{G}=e^{-2kL}[/tex]
[tex]ln(\frac{T}{G})=ln(e^{-2kL})[/tex]
[tex]ln(\frac{T}{G})=-2kL[/tex]
[tex]L=\frac{ln(\frac{T}{G})}{-2k}[/tex]
And replacing values for T, G and k, we can find the width of the barrier:
[tex]L=\frac{ln(\frac{0.03}{3})}{-2(1.98*10^{10})}m[/tex]
[tex]L=1.16*10^{-10}m*\frac{1*10^{9}nm}{1m}[/tex]
L=0.116nm
