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

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

Step-by-step 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 =
`(√(2m(U-E))/(h)`

here h is Planck's constant

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

c = 2319130863.06

and proton have hf =
`(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' =
`(√(2m(U-E))/(h)`

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

c' = 1967510340

so

the factor of incerease in transmisson probability is

probability =
`e^(2L(c-c'))`

probability =
`e^{2* 1* 10^(-9) * (351620523.06)}`

factor probability = 2.02029