In the above question the probability the electron will tunnel through the barrier is 0.011%.
An electron volt is the amount of energy required to move a charge equal to 1e⁻ across a potential difference of 1eV.
To calculate the probability the electron will tunnel through the barrier is calculated as -
Energy of electron
E=80eV
=80×1.6×10⁻¹⁹
=128×10⁻¹⁹
Height of the barrier U=100 eV
=100×1.6×10⁻¹⁹J
Thickness L=0.2×10⁻⁹m
Probability T=e⁻²cl
C=√2m(U-E)/h
=√[2×1.67×10⁻²⁷(160-128)×10⁻¹⁹] / 1.055×10⁻³⁴
=10.34×10⁻²³/1.055×10⁻³⁴
=9.8×10¹¹
2CL=2×9.8×10¹¹×0.2×10⁻⁹
=3.92×10²
T=e⁻²cl
= 1/e⁻³⁹²/¹⁰⁰⁰⁰
=0.0198%
=0.011%
Hence ,the probability the electron will tunnel through the barrier is 0.011%.
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