Estimate the weight of a 1000kg car that is accelerating at 3m/s/s

A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar

denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of $\beta$

Using formula of $\beta$

$\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}$

Put the value into the formula

$\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}$

$\beta=8.86\times10^{9}$

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

$T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}$

Put the value into the formula

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}$

$T=5.45\times10^{-4}$

(b). We need to calculate the tunnel probability for width 1.00 nm

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}$

$T=7.74\times10^{-8}$

Hence, The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

6 0

4 years ago

Which is not a way to accelerate an object?

densk [106]

I would definitely think its B....

3 0

3 years ago

Read 2 more answers

A typical electric oven has two separate heating elements: one on top and one on the bottom. The bottom element is used for baki

Hunter-Best [27]

Answer:

Convection and Radiation mechanisms carry most of the heat

Explanation:

This is because Convection proceeds strongy as heated air rises from the hot element while Radiation is also strong, although the material of the cooking pots will how effective it is.

3 0

3 years ago

The board sandwiched between two other boards shown below weighs 87.5 n. if the coefficient of friction between the boards is 0.

IRINA_888 [86]

Refer to the diagram shown below.

W = 87.5 N, the weight of the sandwiched board.
μ = 0.622, the static coefficient of friction.

From the free body diagram of the sandwiched board, obtain
2μF = W
F = W/(2μ) = 87.5/(2*0.622) = 70.34 N

Answer: 70.34 N