Question

A beam of electrons with 100 eV of energy reach a barrier of height 100.5 eV and width 1 pm. If 10 mA of current is reflected off the barrier, what was the incident current of the beam?

Answer 1

The transmission probability of the electrons is given by

$\alpha =\sqrt {\frac {2m(U-E)}{\hbar^2}}$

$\alpha =\sqrt {\frac {2\cdot 9.1\cdot 10^{-31}\cdot (100.5-100)\cdot 1.6\cdot 10^{-19}}{(1.54\cdot 10^{-34})^2}}m^{-1}=2.47\times 10^{9}\: m^{-1}$

$T=\frac {16\cdot 100\cdot 0.5}{100.5^2}e^{-2\cdot 2.47\cdot 10^9\cdot 1\cdot 10^{-12}}= 0.0795$

$N_i=N_t+N_r$ (Number of electrons incident is equal to the sum of the electrons reflected and transmitted.)

$N_i=0.0795N_i+N_r$

$0.9205N_i=N_r$

$N_i=\frac {N_r}{0.9205}=\frac {\frac {10\cdot 10^{-3}}{1.6\cdot 10^{-19}}}{0.9205}=6.78\times 10^{16}\: electrons$ (Incident electrons)

(Incident current of the beam)