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1 year ago

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What is the binding energy (in J/mol or kJ/mol) of an electron in a metal whose threshold frequency for photoelectrons is 2.50 u

1014 /s?

1 answer:

Aleksandr-060686 [28]1 year ago

4 0

Answer:

binding energy is 99771 J/mol

Exlanation:

given data

threshold frequency = 2.50 × $10^{14}$ Hz

solution

we get here binding energy using threshold frequency of the metal that is express as

$E=h\nu_{o}$ ..................1

here E is the energy of electron per atom $E=\frac{x}{N}$ and h is plank constant i.e. $6.626\times10^{-34} Js$ and x is binding energy

and here N is the Avogadro constant = $6.023\times10^{23}$

so E will $\frac{x}{6.023\times10^{23}}$

E = $3.19\times10^{-19} J$

so put value in equation 1 we get

$\frac{x}{6.023\times10^{23}}$ = 2.50 × $10^{14}$ × $6.626\times10^{-34} Js$

solve it we get

x = 99770.99

so binding energy is 99771 J/mol

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You've always wondered about the acceleration of the elevators in the 101 story-tall Empire State Building. One day, while visit

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To develop this problem we will proceed to convert all units previously given to the international system for which we have to:

$140 lb = 63.5 kg \rightarrow 63.5kg (9.8m/s) =622.3 N$

$120 lb = 54.4 kg \rightarrow 54.4kg (9.8m/s)= 533 N$

$170 lb = 77.1 kg \rightarrow 77.1 kg (9.8m/s) =756 N$

PART A ) From the given values the minimum acceleration will be given for 120Lb and maximum acceleration when 170Lb is reached therefore:

$F = 756 - 622.3$

$F = 133.7N$

Through the Newtonian relationship of the Force we have to:

$F= ma$

$a = \frac{F}{m}$

$a = \frac{133.7}{63.5}$

$a = 2.1m/s^2$

PART B) For the maximum magnitude of the acceleration downward we have that:

$F = 622.3 - 533$

$F = 89.3N$

Through the Newtonian relationship of the Force we have to:

$F= ma$

$a = \frac{F}{m}$

$a = \frac{89.3}{63.5}$

$a = 2.1m/s^2$

$a = 1.04 m/s^2$

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2 years ago

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

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1) At x = 6.6 cm, $E_x=3.47\cdot 10^6 N/C$

2) At x = 6.6 cm, $E_y=0$

3) At x = 1.45 cm, $E_x=-3.76\cdot 10^6N/C$

4) At x = 1.45 cm, $E_y=0$

5) Surface charge density at b = 4 cm: $+62.75 \mu C/m^2$

6) At x = 3.34 cm, the x-component of the electric field is zero

7) Surface charge density at a = 2.9 cm: $+65.25 \mu C/m^2$

8) None of these regions

Explanation:

1)

The electric field of an infinite sheet of charge is perpendicular to the sheet:

$E=\frac{\sigma}{2\epsilon_0}$

where

$\sigma$ is the surface charge density

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

The field produced by a thick slab, outside the slab itself, is the same as an infinite sheet.

So, the electric field at x = 6.6 cm (which is on the right of both the sheet and the slab) is the superposition of the fields produced by the sheet and by the slab:

$E=E_1+E_2=\frac{\sigma_1}{2\epsilon_0}+\frac{\sigma_2}{2\epsilon_0}$

where

$\sigma_1=-2.5\mu C/m^2 = -2.5\cdot 10^{-6}C/m^2\\\sigma_2=64 \muC/m^2 = 64\cdot 10^{-6}C/m^2$

The field of the sheet is to the left (negative charge, inward field), while the field of the slab is the right (positive charge, outward field).

So,

$E=\frac{1}{2\epsilon_0}(\sigma_1+\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}+64\cdot 10^{-6})=3.47\cdot 10^6 N/C$

And the negative sign indicates that the direction is to the right.

2)

We note that the field produced both by the sheet and by the slab is perpendicular to the sheet and the slab: so it is directed along the x-direction (no component along the y-direction).

So the total field along the y-direction is zero.

This is a consequence of the fact that both the sheet and the slab are infinite along the y-axis. This means that if we take a random point along the x-axis, the y-component of the field generated by an element of surface dS of the sheet (or the slab), $dE_y$ , is equal and opposite to the y-component of the field generated by an element of surface dS of the sheet located at exactly on the opposite side with respect to the x-axis, $-dE_y$ . Therefore, the net field along the y-direction is always zero.

3)

Here it is similar to part 1), but this time the point is located at

x = 1.45 cm

so between the sheet and the slab. This means that both the fields of the sheet and of the slab are to the left, because the slab is negatively charged (so the field is outward). Therefore, the total field is

$E=E_1-E_2$

Substituting the same expressions of part 1), we find

$E=\frac{1}{2\epsilon_0}(\sigma_1-\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}-64\cdot 10^{-6})=-3.76\cdot 10^6N/C$

where the negative sign indicates that the direction is to the left.

4)

This part is similar to part 2). Since the field is always perpendicular to the slab and the sheet, it has no component along the y-axis, therefore the y-component of the electric field is zero.

5)

Here we note that the slab is conductive: this means that the charges in the slab are free to move.

We note that the net charge on the slab is positive: this means that there is an excess of positive charge overall. Also, since the sheet (on the left of the slab) is negatively charged, the positive charges migrate to the left end of the slab (at a = 2.9 cm) while the negative charges migrate to the right end (at b = 4 cm).

The net charge per unit area of the slab is

$\sigma=+64\mu C/m^2$

And this the average of the surface charge density on both sides of the slab, a and b:

$\sigma=\frac{\sigma_a+\sigma_b}{2}$ (1)

Also, the infinite sheet located at x = 0, which has a negative charge $\sigma_1=-2.5\mu C/m^2$ , induces an opposite net charge on the left surface of the slab, so

$\sigma_a-\sigma_b = +2.5 \mu C/m^2$ (2)

Now we have two equations (1) and (2), so we can solve to find the surface charge densities on a and b, and we find:

$\sigma_a = +65.25 \mu C/m^2\\\sigma_b = +62.75 \mu C/m^2$

6)

Here we want to calculate the value of the x-component of the electric field at

x = 3.34 cm

We notice that this point is located inside the slab, because its edges are at

a = 2.9 cm

b = 4.0 cm

But slab is conducting , and the electric field inside a conductor is always zero (because the charges are in equilibrium): therefore, this means that the x-component of the electric field inside the slab is zero

7)

We calculated the value of the charge per unit area on the surface of the slab at x = a = 2.9 cm in part 5), and it is $\sigma_a = +65.25 \mu C/m^2$

8)

As we said in part 6), the electric field inside a conductor is always zero. Since the slab in this problem is conducting, this means that the electric field inside the slab is zero: therefore, the regions where the field is zero is

2.9 cm < x < 4 cm

So the correct answer is

"none of these region"

Learn more about electric fields:

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