Question

A scanning tunneling microscope is used to measure small changes in height of a surface by detecting changes in the tunneling current between the tip and the surface. The current is proportional to the tunneling coefficient (ie. I = const * T) which follows the general equation for tunneling through a square well , with C dependent on the molecule and L is the distance. The transmission coefficient at one point is T = 0.01, what is the relative current if the distance is increased from L to 5L? We are interested in I(5L)/I(L). (Note: there may be more information provided than you need to solve the problem.)

Answer 1

`\mathbf{(I(5l))/(I(l) ) =10^(-8)}`

Step-by-step explanation:

We are being told that the current is proportional to the tunneling coefficient
`I(l) = I_0 e^(-2kl)` ;

where l = distance between the tip and the surface.

Let
`I(l) = I_0 e^(-2kl)` ------------ equation (1)

and
`I(5l) = I_0 e^(-2k(5l))` ------------ equation (2)

Dividing equation (2) by (1); we have :

`(I(5l))/(I(l) ) = (I_0 e^(-2k(5l)))/( I_0 e^(-2kl))`

`(I(5l))/(I(l) ) =e^(-2k(5-1)l)`

`(I(5l))/(I(l) ) =(e^(-2kl))^4`

where ;

`(e^(-2kl))` represents the transmission coefficient T = 0.01

Thus; replacing the value for 0.01;we have;

`(I(5l))/(I(l) ) =0.01^4`

`\mathbf{(I(5l))/(I(l) ) =10^(-8)}`