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4 years ago

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The work function (energy needed to remove an electron) of gold is 5.1 eV. Two pieces of gold (at the same potential) are separa

ted by a distance L.For what value of L will the transmission probability for an electron to cross from one to the other be T ≈ 10-3? Assume that G = 1 in the formula for the tunneling probability.

1 answer:

scoundrel [369]4 years ago

7 0

Answer:

The value of L is 0.3 nm.

Explanation:

Given that,

Energy $\phi= 5.1 eV$

Transmission probability = 10⁻³

We need to calculate the value of L

We know that,

Formula of tunneling probability

$T=Ge^{-2kL}$ ....(I)

Where,

$k=\dfrac{\sqrt{2m(E-U)}}{\hbar}$ ....(II)

Where, m = mass of electron

$E = \phi+U$

$\phi = E-U$

Put the value in equation (II)

$k=\dfrac{\sqrt{2\times9.1\times10^{-31}\times5.1\times1.6\times10^{-19}}}{1.055\times10^{-34}}$

$k=1.155\times10^{10} m^{-1}$

From equation (I)

$ln T=-2kLG$

$L=\dfrac{ln T}{-2kG}$

$L=\dfrac{ln 10^{-3}}{-2\times1.155\times10^{10}\times1}$

$L=2.99\times10^{-10}\ m$

$L=0.299\times10^{-9}\ m$

$L=0.3\ nm$

Hence, The value of L is 0.3 nm.

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A person hears the echo of his own voice from a distant hill after 2 seconds. How far away is the person from the hill, if the s

olga nikolaevna [1]

Answer: 330 m

Explanation:

The speed of sound $V$ is defined as the distance traveled by the sound wave $d$ in a especific time $t$ :

$V=\frac{d}{t}$

Where:

$V=330 m/s$ is the speed of sound

$t=\frac{2 s}{2}=1 s$ is half the time the sound wave travels since the person speaks, the sound wave hits the hill and then returns to the person again as an echo

$d$ is the distance between the person and the hill

So, we have to find $d$ :

$d=Vt$

$d=(330 m/s)(1 s)$

Finally:

$d=330 m$

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3 years ago

Speedy Sue, driving at 35.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s

steposvetlana [31]

Answer:

Hence, there will be a collision

Explanation:

First we calculate total distance covered by the speedy sue's car before coming to rest:

2as = Vf² - Vi²

where,

a = deceleration = - 1.9 m/s²

s = distance covered = ?

Vf = Final Velocity = 0 m/s (since car finally stops)

Vi = Initial Velocity = 35 m/s

Therefore,

2(-1.9 m/s²)s = (0 m/s)² - (35 m/s)²

s = 322.37 m

Now, we calculate time taken by car to stop:

Vf = Vi + at

0 m/s = 35 m/s + (-1.9 m/s²)t

t = 18.42 s

Now, we calculate distance traveled by van in this time:

s₁ = V₁t

where,

s₁ = distance traveled by van = ?

V₁ = speed of van = 5.2 m/s

Therefore,

s₁ = (5.2 m/s)(18.42 s)

s₁ = 95.78 m

Now, for collision to occur, the following relation must be satisfied:

s ≥ 160 m + s₁

using values:

322.37 m > 160 m + 95.78 m

322.37 m > 255.78 m

<u>Hence, there will be a collision</u>

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4 years ago

A certain airport runway of length L allows planes to

PilotLPTM [1.2K]

The length of the new runway In terms of (L) is 4L.

<em />

<em>Your question is incomplete, it seems your question is missing the following information below;</em>

"In terms of (L) , what must be the length of the new runway?"

The given parameters;

length of the runway, = L

let the take off speed = V₁

For the newly designed planes;

the take off speed is double of the original = 2V₁

the acceleration for both take is constant = a

The take off speed for the given constant acceleration is given as;

$v^2 = u^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\a = \frac{v^2}{2s} \\\\$

The length of the new runway In terms of (L) is calculated as follows;

$a = \frac{v_1^2}{2s_1} = \frac{v_2^2}{2s_2} \\\\substitute \ the \ corresponding \ values \ for \ initial \ acceleration \ and \ new \\acceleration\\\\\frac{v_1^2}{2L} = \frac{(2v_1)^2}{2s_2}\\\\2s_2v_1^2 = 2L(4v_1^2)\\\\s_2 = \frac{2L(4v_1^2)}{2v_1^2} = \frac{8Lv_1^2}{2v_1^2} = 4L$

Thus, the length of the new runway In terms of (L) is 4L.

Learn more here: brainly.com/question/15206976

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3 years ago

A circular ride has a radius of 32 m If the time of one revolution of a rider is 0.98 s what is the speed of the rider?

11111nata11111 [884]

Given that,

The radius of a circular path, r = 32 m

The time of one revolution of a rider is 0.98 s.

To find,

The speed of the rider.

Solution,

Let v is the speed of the rider. Speed is equal to total distance divided by time taken.

$v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 32}{0.98}\\\\v=205.16\ m/s$

So, the speed of the rider is 205.16 m/s.

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3 years ago

Three 10-cm-long rods form an equilateral triangle. Two of the rods are charged to + 19 nC, the third to - 19 nC.What is the ele

Sveta_85 [38]

<span>We calculate the electric field as follows:
r = </span>√<span>(3)/6 x 19 cm = .05484 m
The angle for the triangle would be 30 on each side.
tan(30) = r/(L/2)

E' = kQ/{r*sqrt[(L/2)^2 + r^2]} = (8.99e9 x 15e-9) / {.05484 * sqrt[(.19/2)^2 + .003]}
</span>E' <span>= 22413 N/C
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|E'| = 22413 N/C
E = 2|E'|sin(30) + |E'| = 49000 N/C</span>

5 0

3 years ago