Question

An electron has a kinetic energy of 10.1 eV. The electron is incident upon a rectangular barrier of height 18.2 eV and width 1.00 nm. If the electron absorbed all the energy of a photon of green light (with wavelength 546 nm) at the instant it reached the barrier, by what factor would the electron's probability of tunneling through the barrier increase

Answer 1

Answer:

factor that the electron's probability of tunneling through the barrier increase 2.02029

Explanation:

given data

kinetic energy = 10.1 eV

height = 18.2 eV

width = 1.00 nm

wavelength = 546 nm

solution

we know that probability of tunneling is express as

probability of tunneling = $e^{-2CL}$   .................1

here C is = $\frac{\sqrt{2m(U-E}}{h}$

here h is Planck's constant

c = $\frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-10.1) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}$  

c = 2319130863.06

and proton have hf = $\frac{hc}{\lambda } = {1240}{546}$ = 2.27 ev

so electron K.E = 10.1 + 2.27

KE = 12.37 eV

so decay coefficient inside barrier is

c' = $\frac{\sqrt{2m(U-E}}{h}$

c' = $\frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-12.37) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}$  

c' = 1967510340

so

the factor of incerease in transmisson probability is

probability = $e^{2L(c-c')}$

probability = $e^{2\times 1\times 10^{-9} \times (351620523.06)}$

factor probability = 2.02029