Answer:
[tex]\mathbf{\dfrac{I(5l)}{I(l) } =10^{-8}}[/tex]
Explanation:
We are being told that the current is proportional to the tunneling coefficient[tex]I(l) = I_0 e^{-2kl}[/tex] ;
where l = distance between the tip and the surface.
Let [tex]I(l) = I_0 e^{-2kl}[/tex] ------------ equation (1)
and [tex]I(5l) = I_0 e^{-2k(5l)}[/tex] ------------ equation (2)
Dividing equation (2) by (1); we have :
[tex]\dfrac{I(5l)}{I(l) } = \dfrac{I_0 e^{-2k(5l)}}{ I_0 e^{-2kl}}[/tex]
[tex]\dfrac{I(5l)}{I(l) } =e^{-2k(5-1)l}[/tex]
[tex]\dfrac{I(5l)}{I(l) } =(e^{-2kl})^4[/tex]
where ;
[tex](e^{-2kl})[/tex] represents the transmission coefficient T = 0.01
Thus; replacing the value for 0.01;we have;
[tex]\dfrac{I(5l)}{I(l) } =0.01^4[/tex]
[tex]\mathbf{\dfrac{I(5l)}{I(l) } =10^{-8}}[/tex]
