Question

What fraction of a beam of $\mathbf{50}\mathbf{}\mathit{e}\mathit{V}\mathbf{}$electrons would get through a $\mathbf{200}\mathbf{}\mathit{V}\mathbf{}\mathbf{1}\mathbf{}\mathit{n}\mathit{m}$ wide electrostatic barrier?

Short answer

The required answer is $1.1\times {10}^{-54}$

Step-by-step solution

Definition of Tunneling

Tunneling defines the penetration of a barrier of high energy by a low-energy wave or particle. For a wide barrier that transmits ineffectively:

$T=16\frac{E}{U}(1-\frac{E}{U})e^{-2kl}$

Where,$k=\frac{\sqrt{2m\left(U-E\right)}}{\bcancel{}}$and l is width of penetration barrier,$U=qV$

Given/known parameters

$V=200V,E=50eV,I=1nm={10}^{-9}m$

Solution

$U=qv=\frac{1.6\times {10}^{-19}\times 200J}{1.6\times {10}^{-19}eV/J}=200eV$

$k=\frac{\sqrt{2\times 9.1\times {10}^{-31}\times (200-50)\times 1.6\times {10}^{-19}J/eV}}{1.05\times {10}^{-34}}$

$k=62.7\times {10}^9{m}^{-1}$

Now,$T=16\left(\frac{50}{200}\right)\left(1-\frac{50}{200}\right)e^{-2\times 62.7\times {10}^9\times {10}^{-9}}=1.1\times {10}^{-54}$

Explanation and Conclusion

The fraction of beam of $50eV$transmitted through a barrier of $200V$and $1nm$width is$1.1\times {10}^{-54}$.