120k views
2 votes
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.

1 Answer

1 vote

Answer:

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.

User Lisa Anne
by
6.5k points