For a step potential function, assume that the potential barrier at x = 0 is twice as large as the total energy of an incident electron, whose velocity is 1 m/s at x < 0. the mass of electron is 9.11×10-31 kg. (a). calculate the probability of finding the electron at a distance d compared with x = 0, where d is 10-10 m into the potential barrier. (b). calculate the penetration depth of the electron impinging on the potential

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

Probability = P(T) = 0.0142

Explanation:

It is given that the potential barrier is 4.70 eV and kinetic energy is 2.80 eV

and the barrier width is 0.40 nm

The probability of tunneling can be found using the equation below

P(T) = 16E/U₀( 1 - E/U₀)e^(-2βL)

β = √2m/h²(U₀ - KE)

Ehere m and h are constants with values m = 511 keV/c² and h = 0.1973 keV.nm/c

so

2m/h² = 2*(511)/(0.1973)²

2m/h² = 26.254 eV.nm²

β = √26.254 (4.70 - 2.80)

β = 7.06/nm

P(T) = 16E/U₀( 1 - E/U₀)e^(-2βL)

P(T) = 16*2.80/4.70( 1 - 2.80/4.70)e^(-2*7.06*0.40)

P(T) = 9.532*(0.404)e^(-5.6)

P(T) = 0.0142

Therefore, the probability that the electron will tunnel through the potential barrier of height 4.70 eV with kinetic energy of 2.80 eV and barrier width of 0.40 nm is 0.0142.

[tex]2.9\times 10^{-10}\ m[/tex]

Explanation:

E = Initial kinetic energy = 5.1 eV

[tex]u_0[/tex] = Potential barrier height = 10 eV

P = Probability = [tex]\dfrac{0.6}{100}=0.6\times 10^{-2}[/tex]

m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]

Constant A is given by

[tex]A=16\dfrac{E}{u_0}(1-\dfrac{E}{u_0})\\\Rightarrow A=16\times \dfrac{5.1}{10}(1-\dfrac{5.1}{10})\\\Rightarrow A=3.9984[/tex]

K is given by

[tex]K=\dfrac{\sqrt{2m(u_0-E)}}{\hslash}\\\Rightarrow K=\dfrac{\sqrt{2\times 9.1\times 10^{-31}(10-5.1)\times 1.6\times 10^{-19}}}{1.055\times 10^{-34}}\\\Rightarrow K=11322472269.49086\ /m[/tex]

We have the relation

[tex]P=Ae^{-2Kl}\\\Rightarrow \dfrac{P}{A}=e^{-2Kl}\\\Rightarrow \dfrac{A}{P}=e^{2Kl}\\\Rightarrow ln\dfrac{A}{P}=2Kl\\\Rightarrow l=\dfrac{ln\dfrac{A}{P}}{2K}\\\Rightarrow l=\dfrac{ln\dfrac{3.9984}{0.6\times 10^{-2}}}{2\times 11322472269.49086}\\\Rightarrow l=2.87123\times 10^{-10}\ m[/tex]

The width of the barrier is [tex]2.9\times 10^{-10}\ m[/tex]

Sustainable technologies currently release more greenhouse gases than nonrenewable energy sources release.

Explanation:I took the test

Question:

An electron with kinetic energy 2.80 eV encounters a potential barrier of height 4.70 eV. If the barrier width is 0.40 nm, what is the probability that the electron will tunnel through the barrier?

a) 3.5 × 10-3

b) 7.3 × 10-3

c) 1.4 × 10-2

d) 3.5 × 10-2

e) 2.9 × 10-3

Given Information:

Potential barrier = U₀ = 4.70 eV

kinetic energy = E = 2.80 eV

Barrier width = L = 0.40 nm

Required Information:

Probability of tunneling = ?

c) Probability of tunneling = 1.42×10⁻²

Explanation:

The probability of tunneling is given by

T(L,E) = 16E/U₀( 1 - E/U₀)e^(-2βL)

Where β is given by

β = √2m/h²(U₀ - KE)

Where m is the mass of electron and h is Planck’s constant

m = 511 keV/c² and h = 0.1973 keV.nm/c

2m/h² = 2*(511)/(0.1973)²

2m/h² = 26.254 eV.nm²

β = √26.254 (4.70 - 2.80)

β = 7.06/nm

T(L,E) = 16E/U₀( 1 - E/U₀)e^(-2βL)

T(L,E) = 16*2.80/4.70( 1 - 2.80/4.70)e^(-2*7.06*0.40)

T(L,E) = 9.532*(0.404)e^(-5.6)

T(L,E) = 0.0142

or

T(L,E) = 1.42×10⁻²

36

Explanation:

The correct answer is option B.

Explanation:

If a producer has patent for a product it means others cannot produce it. So, this a barrier to entry.

Similarly, in case of licensing only the producer holding license can produce. So, it is barrier to entry.

The control to a crucial input will also restrict others to produce. So, this will also be considered as a barrier.

Economies of scale is cost advantage to a firm because of its large size. It is a barrier to entry as it would discourage other new firms to enter.

Absence of economies of scale is not a barrier to entry in the product market.