Question #174858
A electric beam is incident on a potential barrier at height 0.03ev and of infinite with find the reflection coefficient of the barrier if the energy of the incident electric beam be (a)0.04ev(b)0.025ev(c)0.03ev.


1
Expert's answer
2021-03-24T19:58:18-0400

GIVEN:

Vo=0.03eVa)E1=0.04b)E2=0.025c)E3=0.03V_o=0.03eV \\a) E_1=0.04\\b)E_2=0.025\\c)E_3=0.03

first we find the transmission coefficient (T) for the step like potential:

T=4EEVo(E+EVo)2\boxed{T={4\sqrt E \sqrt{E-V_o}\over (\sqrt E + \sqrt{E-V_o})^2}} ............1


then reflection coefficient (R).

R=1T\boxed{R=1-T} ...............................2


now using 1 and 2 we calculate T1,T2,T3T_1,T_2,T_3 AND R1,R2,R3R_1,R_2,R_3

a)

T1=40.040.040.03(0.04+0.040.03)2=0.889\boxed{T_1={4\sqrt 0.04 \sqrt{0.04-0.03}\over (\sqrt0.04 + \sqrt{0.04-0.03})^2}=0.889} \\


R1=1T=0.111\boxed{R_1=1-T=0.111}

b)

T2=40.0250.0250.03(0.025+0.0250.03)2=error\boxed{T_2={4\sqrt 0.025 \sqrt{0.025-0.03}\over (\sqrt0.025 + \sqrt{0.025-0.03})^2}=error} \\ no value exists because ,E<V and squre root is not defined for negative value.

therefore, value of R2=R_2= not defined


c)

T3=40.030.030.03(0.03+0.030.03)2=0\boxed{T_3={4\sqrt 0.03 \sqrt{0.03-0.03}\over (\sqrt0.03 + \sqrt{0.03-0.03})^2}=0}


therefore

R3=10=1\boxed{R_3=1-0=1}


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