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Chapter 6: Q54P (page 1357)

An electron with initial kinetic energy 5.5 eV encounters a square potential barrier of height 10.0 eV. What is the width of the barrier if the electron has a 0.50% probability of tunnelling through the barrier?

Short Answer

Expert verified

The magnitude of the barrier width is L= 0.31 nm .

Step by step solution

01

(a) Identification of the concept.

The tunnelling probability is given as,

T=Ge2kL

Here, T is the tunnelling probability, G=16EU01-EU0 .

Also, k=2mU0-Ehis the constant.

02

(b) Determination of the magnitude of the barrier width.

The given values are,

E=5.5eVU0=10.0eVm=9.11脳10-31kgT=0.0050

Substitute all the values to determine k,

k=29.11脳10-31kg4.5eV1.60脳10-19J/eV1.054脳10-34J.s

Substitute all the values to determine G,

G=165.5eV10.0eV1-5.5eV10.0eV=3.96

From equation (i), solve for the barrier width,

L=-12kInTG=121.09脳1010m-1In0.00503.96=3.1脳10-10m=0.31nm Thus, the barrier width magnitude is L= 0.31 nm

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