# Thread: Question on potential calculation

1. ## Question on potential calculation

Hi everyone
need help for this question, im not sure what formula to use.really appreciate if someone can help me.

A proton with an energy of 1.5 eV impinging on a potential barrier with
V= 15 eV and a width of 2Å . Assume that the proton mass is equal to the
rest mass of electron. Calculate the probability of the proton tunnelling
through the potential barrier

2. Originally Posted by anderson
Hi everyone
need help for this question, im not sure what formula to use.really appreciate if someone can help me.

A proton with an energy of 1.5 eV impinging on a potential barrier with
V= 15 eV and a width of 2Å . Assume that the proton mass is equal to the
rest mass of electron. Calculate the probability of the proton tunnelling
through the potential barrier

[This isn't only Math, well it's the Math of Quantum Physics]

The probability of tunnelling is given (approximately) by $P \approx e^{-2\alpha L}$, where $\alpha = \frac{\sqrt{2m(U_0-E)}}{h(2\pi)^{-1}}$.

The key is to multiply the whole fraction by $\frac{c}{c}$, (inside the numerator's square root, you should multiply by $c^2$). The reason for this is that, now, you get hc (a known constant) in the denominator and you get $mc^2$ in the numerator, which is just energy $[E=mc^2]$.

$\alpha = \frac{\sqrt{2m(U_0-E)}}{h(2\pi)^{-1}}=\frac{\sqrt{2mc^2(U_0-E)}}{hc(2\pi)^{-1}}$.

Now simply substitute everything in, note that the constant hc is roughly 197eVnm.

The rest mass (m) is approximately $5\cdot 10^5 eVc^{-2}$

Multiplying both sides by $c^2$, $mc^2$, which is energy, just becomes 5*10^5 eV.

Substituting everything, alpha becomes:

$\alpha = \frac{\sqrt{2(5\cdot10^5eV)(13.5eV)}}{197eVnm} \approx 18.65 nm^{-1}$

The barrier width is 2 angstrom which is 0.2nm:

$P \approx e^{-2\alpha L} \approx e^{-2(18.65nm^{-1})(0.2nm)}$

$P \approx e^{-7.46}$

And, there you go

Hope I helped you, anderson.

3. Dear mathemagister

thank you so much, relly helped me.

thank you again.

4. Originally Posted by anderson
Dear mathemagister

thank you so much, relly helped me.

thank you again.