Question

The work function (energy needed to remove an electron) of gold is 5.1 eV. Two pieces of gold (at the same potential) are separated by a distance L.For what value of L will the transmission probability for an electron to cross from one to the other be T ≈ 10-3? Assume that G = 1 in the formula for the tunneling probability.

Answer 1

Answer:

The value of L is 0.3 nm.

Explanation:

Given that,

Energy $\phi= 5.1 eV$

Transmission probability = 10⁻³

We need to calculate the value of L

We know that,

Formula of tunneling probability

$T=Ge^{-2kL}$....(I)

Where,

$k=\dfrac{\sqrt{2m(E-U)}}{\hbar}$....(II)

Where, m = mass of electron

$E = \phi+U$

$\phi = E-U$

Put the value in equation (II)

$k=\dfrac{\sqrt{2\times9.1\times10^{-31}\times5.1\times1.6\times10^{-19}}}{1.055\times10^{-34}}$

$k=1.155\times10^{10} m^{-1}$

From equation (I)

$ln T=-2kLG$

$L=\dfrac{ln T}{-2kG}$

$L=\dfrac{ln 10^{-3}}{-2\times1.155\times10^{10}\times1}$

$L=2.99\times10^{-10}\ m$

$L=0.299\times10^{-9}\ m$

$L=0.3\ nm$

Hence, The value of L is 0.3 nm.