Transmission probability

The transmission probability for a particle of energy $E$ on an abrupt step of height $U_0$ when $E>U_0$ is $T=4k_1k_2/(k_1+k_2)^2$ where $k_1$ is the wave vector when $U=0$ and $k_2$ is the wave vector when $U=U_0$. For $U_0= 1eV$ find energy $E$ for an electron transmission probability of $T=0.99$. The same for $T=0.01$

$k_{1,2}$ are wave numbers and thus their definition is $k=2π/λ=2π/(h/p)=p/\hbar$

In the free region the energy of particle is E, and the wave function is a wave. Therefore


Above the step the wave function is still a wave with another k:


$T=(4*k_1 k_2)/(k_1+k_2 )^2$

If $T=0.99$ for $U =1 eV$ then

$0.99= \frac{8m\sqrt{E(E-U)}}{2mE+2m(E-U)+4m\sqrt{E(E-U)}}$

$0.99= \frac{4}{\sqrt{E/(E-U)}+\sqrt{(E-U)/E}+2}$

$0.99= 4/[x+(1/x)+2]$

$0.99x+0.99/x+1.98=4$ that is $0.99x^2-2.02*x+0.99=0$  or $x=1.2222$..

$E/(E-1)=1.2222…$  that is $E=5.5 eV$

If $T =0.01$ and $U =1 eV$ then:

$0.01= 4/[x+(1/x)+2]$   that is $0.01x+0.01/x+0.02=4$ or $0.01x^2-3.98*x+0.01=0$  or $x=397.997$

$E/(E-1)=397.9975$   that is $E=397.9975/396.9975=1.00252 eV$