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

The value of L is 0.3 nm.

Step-by-step 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=(√(2m(E-U)))/(\hbar)`....(II)

Where, m = mass of electron

`E = \phi+U`

`\phi = E-U`

Put the value in equation (II)

`k=\frac{\sqrt{2*9.1*10^(-31)*5.1*1.6*10^(-19)}}{1.055*10^(-34)}`

`k=1.155*10^(10) m^(-1)`

From equation (I)

`ln T=-2kLG`

`L=(ln T)/(-2kG)`

`L=(ln 10^(-3))/(-2*1.155*10^(10)*1)`

`L=2.99*10^(-10)\ m`

`L=0.299*10^(-9)\ m`

`L=0.3\ nm`

Hence, The value of L is 0.3 nm.