Home Page Textbooks Physics Page 863: Lesson Check

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 863: Lesson Check

Exercise 14

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The energy distribution in a blackbody is independent of its material – it depends only on the temperature. Hence, the steel and plywood would have the same energy radiated

Result

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Steel and plywood would have the same energy radiated

Exercise 15

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Quantized energy means that bodies can have only certain discrete values of energy and energy exchange between them also occurs in there increments that are fixed (total energy of the body is a positive integer times some $E_0$ called quantum of energy). In the continuous specter on the other hand energy can have any value and can change for any amount in principle.

Result

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Exercise 16

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Energy of a photon is uniquely determined by its’ frequency (OR its’ wavelength) since $E=hnu$ (OR $E=hc/lambda$).

Result

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Exercise 17

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To explain the spectrum of a black body using classical physics, 19th century physicists got the distribution law that predicted that as frequencies grow so does the abundance of photons of that frequency (there are more and more photons as you move from infrared towards ultraviolet part of the spectrum). This is not physically admissible since then the body would radiate infinite energy and this was called ultraviolet catastrophe.

Result

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Exercise 18

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The photon of red light has lower frequency than a photon of blue light and thus always has less energy. But if there are more photons in a beam of red light than photons in a beam of blue light than it is possible for the red beam to deliver more energy.

Result

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Exercise 19

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Because this proves that light isn’t absorbed continuously by the electron but only single photons are absorbed. If the photon is not energetic enough to eject the electron from the surface (to overcome the materials work function) then no electrons would be ejected as it has been proved in experiments.

Result

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Exercise 20

Solution 1

Solution 2

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$textbf{Given:}$

$T = 5800$ K

$textbf{Find:} f_{peak}$

We can compute for $f_{peak}$ using Wien’s displacement law as follows:

$$
begin{align*}
f_{peak} &= (5.88 times 10^{10} text{ Hz}) times 5800 text{ K} \
&= boxed{3.4 times 10^{14} text{Hz}}
end{align*}
$$

Result

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$$
f_{peak} = 3.4 times 10^{14} text{Hz}
$$

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Using Wien’s law we directly calculate

$$
nu_{peak} = 5.88times10^{10}text{ Hz}times 5800text{ K} =3.4times 10^{14}text{ Hz}.
$$

Result

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Exercise 21

Solution 1

Solution 2

Step 1

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$textbf{Given:}$

$E = 1.2$ J

$v = 0.82$ Hz

$textbf{Find}$ quantum number, $n$

We can calculate $n$ using the formula for quantized energy such that:

$$
begin{align*}
E &= nhf \
n &= dfrac{E}{hf} \
&= dfrac{1.2 text{ J}}{6.626 times 10^{-34} text{ Js} times 0.82 text{ Hz}} \
&= boxed{2.2 times 10^{33}}
end{align*}
$$

Result

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$$
n = 2.2 times 10^{33}
$$

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If we suppose quantized energy it has to be of the form of $E=nhnu$ which yields for the quantum number $n$

$$
n=frac{E}{hnu}=frac{1.2text{ J}}{0.82text{ Hz}times6.62times10^{-34}text{ J s}} =2.2times 10^{33}.
$$

Result

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Exercise 22

Solution 1

Solution 2

Step 1

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$textbf{Given: }$

$E_1 = 4.2 times 10^{-19} text{ J}$

$E_2 = 1.9 text{ eV} times left(dfrac{1.6 times 10^{-19} text{ J}}{1 text{ eV}}right) = 3.04 times 10^{-19} text{ J}$

To calculate for the frequencies, we use the following formula:

$$
begin{gather}
E = hf
end{gather}
$$

a) Using Eq (1), we can calculate for $f_1$ as follows:

$$
begin{align*}
f_1 &= dfrac{E}{h} \
&= dfrac{4.2 times 10^{-19} text{ J}}{6.626 times 10^{-34} text{ Js}} \
&= boxed{6.3 times 10^{14} text{ Hz}}
end{align*}
$$

Step 2

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b) Using Eq (1), we can calculate for $f_2$ as follows:

$$
begin{align*}
f_2 &= dfrac{E}{h} \
&= dfrac{3.04 times 10^{-19} text{ J}}{6.626 times 10^{-34} text{ Js}} \
&= boxed{4.6 times 10^{14} text{ Hz}}
end{align*}
$$

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a) $6.3 times 10^{14}$ Hz

b) $4.6 times 10^{14}$ Hz

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a) From the relation $E=hnu$ we get for the frequency

$$
nu = frac{E}{h}=frac{4.2times10^{-19}text{ J}}{6.62times 10^{-34}text{ J s}}=6.3times10^{14}text{ Hz}.
$$

b)
$$
nu = frac{E}{h} = frac{1.9text{ eV}}{6.62times10^{-34}text{ J s}} = frac{1.9times1.6times 10^{-19}text{ J}}{6.62times 10^{-34}text{ J s}} = 4.6times10^{14}text{ Hz}.
$$

Result

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Exercise 23

Solution 1

Solution 2

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$textbf{Given:}$

$f = 9.3 times 10^{14}$ Hz

$W_o = 3.6 text{ eV} times left(dfrac{1.6 times 10^{-19} text{ J}}{1 text{ eV}}right) =5.76 times 10^{-19} text{ J}$

$textbf{Find:} KE_{max}$

To calculate for $KE_{max}$, we use the Einstein’s equation for the photoelectric effect as follows:

$$
begin{align*}
E &= hf – W_o \
KE_{max} &= hf – W_o \
&= (6.626 times 10^{-34} text{ Js} times 9.3 times 10^{14} text{ Hz}) – 5.76 times 10^{-19} text{ J} \
&= boxed{4.02 times 10^{-20} text{ J}}
end{align*}
$$

Result

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$$
KE_{max} = 4.02 times 10^{-20} text{ J}
$$

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From Einstein’s formula for photoelectric effect we get

$$
hnu = W_0 + E_k.
$$
This yields for maximum kinetic energy

$$
E_k = hnu – W_0 = 6.62times10^{-34}text{ J s}times 9.3times 10^{14}text{ Hz}- 3.6times1.6times10^{-19}text{ J} = 0.4times 10^{-19}text{ J}.
$$

Result

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