Question

. A 0.100-mA electron beam with kinetic energy 54.0 eV enters a sharply defined region of lower potential where the kinetic energy of the electrons is increased by 10.0 eV. What current is reflected at the boundary? (This simulates electron scattering at normal incidence from a metal surface, as in the Davisson–Germer experiment.)

Answer 1

Step-by-step explanation:

The reflection coefficient is the ratio of reflected flux to the incident flux

The reflection coefficient of a srep potential when a particle is
`E(> U(x))` is incident on the barrier is given as:

`R=((K_(1)-K_(2))/(K_(1)+K_(2)) )^2`

Where
`k_(1)` is the propagation constant of the incident wave and
`k_(1)` is the propagation constant of transmitted wave.

The expression for the
`k_(1)` is given as:

`k_(1)=\sqrt{(2mE)/(h^2) } \\\\=\sqrt{(2(mc^2)E)/(h^2c^2) } \\\\=(√(2(mc^2)E))/(hc)`

Here
`m` is the electron mass;
`h` the reduced plancks constant;
`c` the speed of light and
`E` the energy of the incident wave.

substituting
`0.511*10^(6)eV` for
`mc^2`;
`54.0eV` for
`E` and
`1.973*10^(3)eV.A^(o)/c` for
`h` in the equation

`k_(1)=(√(2(mc^2)E))/(hc)\\\\=\frac{\sqrt{2(0.511*10^(6)eV)(54.0eV)}}{1.973*10^(3)eV.A^(o)/c}\\\\(3.765/A^o)((1A^o)/(10^(-10)m))\\\\=3.765*10^(10)/m` this is the propagation constant for the incident wave.

For
`K_(2)`

`K_(2)=\sqrt{(2m(U+E))/(h^2) } \\\\=\sqrt{(2(mc^2)(U+E))/(h^2c^2) }\\\\=(√(2(mc^2)(U+E)))/(hc)`

Here
`m` is the electron mass;
`h` the reduced plancks constant;
`c` the speed of light and
`E` the energy of the incident wave.

substituting
`0.511*10^(6)eV` for
`mc^2`;
`64.0eV` for
`E` and
`1.973*10^(3)eV.A^(o)/c` for
`h` in the equation

`K_(2)=(√(2(mc^2)(U+E)))/(hc)`

`K_(2)=\frac{\sqrt{2(0.511*10^(6)eV)(64.0eV)}}{1.973*10^(3)eV.A^(o)/c}\\\\=(4.1/A^o)((1A^o)/(10^(-10)m))\\\\=4.1*10^(10)/m`this is the propagation constant for the transmitted wave.

substituting
`3.765*10^(10)/m` for
`K_(1)` and
`4.1*10^(10)/m`
`K_(2)`

`R=((3.765*10^(10)/m-4.1*10^(10)/m)/(3.765*10^(10)+4.1*10^(10)/m) )^2\\\\=0.0018`

The ratio of the reflected current to the incident current is the reflection coefficient.

`R=((I_(reflected))/(I_(incident))\\\\I_(reflected)=RI_(incident)`

substituting 0.0018 for R and 0.100mA for
`I_(incident)`

`I_(reflected)=(0.008)(0.100)\\\\=0.18*10^(-16)A`