Probability of penetrating a potential barrier

[EGN123]

Homework Statement

The probability for a particle of energy E<<V₀ to penetrate a potential barrier of height V₀ and width d is approximately \frac{16E}{V_0}exp\left[\frac{-2d\sqrt{2m(V_0-E)}}{\hbar}\right].
An electron moves between two potential barriers of height V₀ and 2v₀ that are of widths 2a and a respectively . For what range of energies is it more likely to exit through the right barrier than through the left?

Homework Equations

Given in question.

The Attempt at a Solution

P_\text{left}=\frac{16E}{V_0}exp\left[\frac{-4a\sqrt{2m(V_0-E)}}{\hbar}\right]

P_\text{left}=\frac{16E}{2V_0}exp\left[\frac{-2a\sqrt{2m(2V_0-E)}}{\hbar}\right]

I have attempted to solve the inequality P_right>P_left. Due to the two square roots I had to square the entire expression twice to obtain an expression without surds, leaving me to solve:
9E^2+\left(\frac{10\hbar^2}{4a^2}(\ln{2})^2-12V_0\right)E+4V_0^2-\frac{3\hbar^2}{a^2}(\ln{2})^2V_0&gt;0

I know I can solve this to find a range of energies, however I think there should be an easier way to solve it, since it is an exam question with limited time.

 

[PeroK]

Homework Statement

The probability for a particle of energy E<<V₀ to penetrate a potential barrier of height V₀ and width d is approximately \frac{16E}{V_0}exp\left[\frac{-2d\sqrt{2m(V_0-E)}}{\hbar}\right].
An electron moves between two potential barriers of height V₀ and 2v₀ that are of widths 2a and a respectively . For what range of energies is it more likely to exit through the right barrier than through the left?

Homework Equations

Given in question.

The Attempt at a Solution

P_\text{left}=\frac{16E}{V_0}exp\left[\frac{-4a\sqrt{2m(V_0-E)}}{\hbar}\right]

P_\text{left}=\frac{16E}{2V_0}exp\left[\frac{-2a\sqrt{2m(2V_0-E)}}{\hbar}\right]

I have attempted to solve the inequality P_right>P_left. Due to the two square roots I had to square the entire expression twice to obtain an expression without surds, leaving me to solve:
9E^2+\left(\frac{10\hbar^2}{4a^2}(\ln{2})^2-12V_0\right)E+4V_0^2-\frac{3\hbar^2}{a^2}(\ln{2})^2V_0&gt;0

I know I can solve this to find a range of energies, however I think there should be an easier way to solve it, since it is an exam question with limited time.

Just looking at the maths - why not just cancel the $\frac{16E}{V_0}$ before you start?

 

[EGN123]

Just looking at the maths - why not just cancel the $\frac{16E}{V_0}$ before you start?

I did that as part of the maths which led to the final inequality, I just didnt show the steps as its just rearranging the original inequality.

 

[PeroK]

I did that as part of the maths which led to the final inequality, I just didnt show the steps as its just rearranging the original inequality.

Yes, of course! The clue is $E << V_0$ I think. That suggests a binomial expansion to me.

 

[EGN123]

Yes, of course! The clue is $E << V_0$ I think. That suggests a binomial expansion to me.

I hadn't thought of that at all. I'll give it a try, thanks!