The probability of tunnelling is given (approximately) by , where .
The key is to multiply the whole fraction by , (inside the numerator's square root, you should multiply by ). The reason for this is that, now, you get hc (a known constant) in the denominator and you get in the numerator, which is just energy .
Now simply substitute everything in, note that the constant hc is roughly 197eVnm.
The rest mass (m) is approximately
Multiplying both sides by , , which is energy, just becomes 5*10^5 eV.
Substituting everything, alpha becomes:
The barrier width is 2 angstrom which is 0.2nm:
And, there you go
Hope I helped you, anderson.