An electron with kinetic energy 2.80 eV encounters a potential barrier of height 4.70 eV. If the barrier width is 0.40 nm, what is the probability that the electron will tunnel through the barrier?

Probability = P(T) = 0.0142

Explanation:

It is given that the potential barrier is 4.70 eV  and kinetic energy is 2.80 eV

and the barrier width is 0.40 nm

The probability of tunneling can be found using the equation below

P(T) = 16E/U₀( 1 - E/U₀)e^(-2βL)

β = √2m/h²(U₀ - KE)

Ehere m and h are constants with values m = 511 keV/c² and h = 0.1973 keV.nm/c

so

2m/h² = 2*(511)/(0.1973)²

2m/h² = 26.254 eV.nm²

β = √26.254 (4.70 - 2.80)

β = 7.06/nm

P(T) = 16E/U₀( 1 - E/U₀)e^(-2βL)

P(T) = 16*2.80/4.70( 1 - 2.80/4.70)e^(-2*7.06*0.40)

P(T) = 9.532*(0.404)e^(-5.6)

P(T) = 0.0142

Therefore, the probability that the electron will tunnel through the potential barrier of height 4.70 eV with kinetic energy of 2.80 eV and barrier width of 0.40 nm is 0.0142.

yes

step-by-step explanation:

yes

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.