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1 year ago

Calculate the lowest energy (in ev) for an electron in an infinite well having a width of 0.050 mm.

Physics

1 answer:

MA_775_DIABLO [31]1 year ago

7 0

The lowest energy of electron in an infinite well is 1.2*10^-33J.

To find the answer, we have to know more about the infinite well.

<h3>What is the lowest energy of electron in an infinite well?</h3>

It is given that, the infinite well having a width of 0.050 mm.
We have the expression for energy of electron in an infinite well as,

$E_n=\frac{n^2h^2}{8mL^2}$

where;

$m=9.1*10^{-31}kg\\L=0.050*10^{-3}m\\h=6.63*10^{-34}Js\\n=1$

Thus, the lowest energy of electron in an infinite well is,

$E_1=\frac{(6.63*10^{-34})^2}{8*9.1*10^{-31}*(0.050*10^{-3})}=1.2*10^{-33}J$

Thus, we can conclude that, the lowest energy of electron in an infinite well is 1.2*10^-33J.

Learn more about the infinite well here:

brainly.com/question/20317353

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<h2>Answer</h2>

The answer to this question is $11.67 s$

<h2>Explanation</h2>

As we know that accelartion is the rate of change of velocity. So, it can be write as

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So, the right answe is $11.67 s$

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3 years ago

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Answer:

Yes

Explanation:

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The radius of a small ball is around 3.79747 cm. The radius of a basketball is about 3.16 times larger. What is the ratio of the

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3 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

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E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

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$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

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For r > r2, E =0 if

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