.. DATA As an intern at a research lab, you study the transmission of electrons through a potential barrier. You know the height of the barrier, 8.0 eV, but must measure the width Lof the barrier. When you measure the tunneling probability Tas a

function of the energy Eof the electron, you get the results shown in the table.

(a) For each value of E, calculate the quantities Gand k that appear in Eq. (40.42). Graph ln T/G versus k. Explain why your

data points, when plotted this way, fall close to a straight line. (b) Use the slope of the best-fit straight line to the data in part (a)

From equation 40.42, the tunnelling probability T that the particle with energy E gets through a potential barrier of height

Where

Therefore the quantity G is

Calculations

(:1) Given that the height of the potential [merrier is {r}. — 8.0 (N and the

mass of an electron is_\ , we can use mutation (2) to get.

the value of (I (:(‘n respules (.0 each value of E.

And. use equation (3] to get the value of 1c corresponding to each mluc of L7.

And, use equation (3) to get thevalueof 1c corresponds each mluc of E.
Next, we get. the value of

we list n“ of these valuesin below.

ThegRaphversus 1; is shown below.
Rearranging equation (1), We get:

As we can see, the quantity In (T / G ) is directly proportional to r3-

This is why we did expect to see a straight line when graphing In (T / G ) versus H.

(b) From equation (4), We can see that graphing 111 (T / G) versus A: should gives us a straight line with slope = —2L-

So, the width of the barrier is»

L = M

—2

Using a best-?t method, we ?nd that the slope of the line is —1.4 X 10—9 m (from the graph)-

Thus,

Therefore the value of L is