Answer:
The value of L is 0.3 nm.
Explanation:
Given that,
Energy [tex]\phi= 5.1 eV[/tex]
Transmission probability = 10⁻³
We need to calculate the value of L
We know that,
Formula of tunneling probability
[tex]T=Ge^{-2kL}[/tex]....(I)
Where,
[tex]k=\dfrac{\sqrt{2m(E-U)}}{\hbar}[/tex]....(II)
Where, m = mass of electron
[tex]E = \phi+U[/tex]
[tex]\phi = E-U[/tex]
Put the value in equation (II)
[tex]k=\dfrac{\sqrt{2\times9.1\times10^{-31}\times5.1\times1.6\times10^{-19}}}{1.055\times10^{-34}}[/tex]
[tex]k=1.155\times10^{10} m^{-1}[/tex]
From equation (I)
[tex]ln T=-2kLG[/tex]
[tex]L=\dfrac{ln T}{-2kG}[/tex]
[tex]L=\dfrac{ln 10^{-3}}{-2\times1.155\times10^{10}\times1}[/tex]
[tex]L=2.99\times10^{-10}\ m[/tex]
[tex]L=0.299\times10^{-9}\ m[/tex]
[tex]L=0.3\ nm[/tex]
Hence, The value of L is 0.3 nm.
