Home Page Textbooks Physics: Principles and Problems Page 730: Practice Problems

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 730: Practice Problems

Exercise 1

Step 1

1 of 2

In order to solve this problem we are going to use the fact that each eV has $1.6times 10^{-19}$ J. So we have that

$$
E=2.3times 1.6times 10^{-19}
$$

Finally, we have

$$
boxed{E=3.68times 10^{-19}textrm{ J}}
$$

Result

2 of 2

$$
E=3.68times 10^{-19}textrm{ J}
$$

Exercise 2

Step 1

1 of 4

$dfrac{1}{2}m_e v_e^2$ = energy in joule

$m_e=9.10938215times 10^{-31}$ Kg
and $v_e=6.2times 10^6$ m/s and $e_{charge}=1.602176487times 10^{-19}$ C

Step 2

2 of 4

$dfrac{1}{2}times 9.10938215 times 10^{-31}times(6.2times 10^6)^2=1.7550823249times 10^{-17}$Joule

to convert from Joule to eV we divide by charge of electron ($e_{charge}$)

Step 3

3 of 4

energy in eV = $dfrac{1.7550823249times 10^{-17}}{1.602176487times 10^{-19} }=$ 109.277802 eV

Result

4 of 4

Energy in eV = 109.277802 eV

Exercise 3

Step 1

1 of 2

In order to solve this problem we have to find the energy in Joules first. We are going to use the fact that each eV has $1.6times 10^{-19}$ J. So we have that

$$
E=2.3times 1.6times 10^{-19}
$$

Finally, we have

$$
E=3.68times 10^{-19}textrm{ J}
$$

Now, the kinetic energy by definition is

$$
E=frac{mv^2}{2}
$$

from where one can express the speed as

$$
v=sqrt{frac{2E}{m}}=sqrt{frac{2times 3.68times 10^{-19}}{9.11times 10^{-31}}}
$$

Finally, we get that

$$
boxed{v=9times 10^5 ,frac{textrm{m}}{textrm{s}}}
$$

Result

2 of 2

$$
v=9times 10^5 , frac{textrm{m}}{textrm{s}}
$$

Exercise 4

Step 1

1 of 2

In order to solve this problem, we have to use the photoelectric effect formula which says that the maximum kinetic energy and the stopping potential are related as

$$
K_{max}=-qV_0
$$

Now, we can transfer our values from Joules to eVs but we can more elegantly note that eV is a unit charge times one Volt which gives that in our case

$$
K_{max}=-(-1e)times 5.7V
$$

Finally

$$
boxed{K_{max}=5.7textrm{ eV}}
$$

Result

2 of 2

$$
K_{max}=5.7textrm{ eV}
$$

Exercise 5

Step 1

1 of 2

In order to solve this problem, we are going to use the relation of the stopping potential and maximal kinetic energy in a photoelectric effect

$$
K_{max}=-qV_0=1.6times 10^{-19}times 3.2
$$

Finally, we get that

$$
boxed{K_{max}=5.12times 10^{-19}textrm{ J}}
$$

Result

2 of 2

$$
K_{max}=5.12times 10^{-19}textrm{ J}
$$

Exercise 6

Step 1

1 of 2

In order to solve this problem we are going to use the fact that the threshold frequency is given as
$f_0=frac{c}{lambda_0}=frac{3times 10^8}{310times 10^{-9}}$
$boxed{f_0=9.7times 10^{-14}textrm{ Hz}}$
Now, the work function is given as

$$
W=hf_0=6.63times 10^{-34}times 9.7times 10^{-14}times 6.24 times 10^{18}
$$

Finally

$$
boxed{W=4textrm{ eV}}
$$

Result

2 of 2

$$
f_0=9.7times 10^{-14}textrm{ Hz}
$$

$$
W=4textrm{ eV}
$$

Exercise 7

Step 1

1 of 2

In order to solve this problem we are going to use the equation of the photoelectric effect which tells us that

$$
K_{max}=E_{ph}-W=frac{hc}{lambda}-hf_0
$$

After we plug in the values we have that

$$
K_{max}=frac{1240times 10^{-9}}{425times 10^{-9}}-1.96=2.92-1.96
$$

Finally, we find that

$$
boxed{K_{max}=0.96textrm{ eV}}
$$

Result

2 of 2

$$
K_{max}=0.96textrm{ eV}
$$

Exercise 8

Step 1

1 of 2

In order to solve this problem, we are going to use the equation of the photoelectric effect which tells us that

$$
K=E_{ph}-W=frac{hc}{lambda}-W
$$

Now, we can express the work function as

$$
W=frac{hc}{lambda}-K=frac{1240times 10^{-9}}{193}-3.5=6.4-3.5
$$

Finally, we have that

$$
boxed{W=2.9textrm{ eV}}
$$

Result

2 of 2

$$
W=2.9textrm{ eV}
$$

Exercise 9

Step 1

1 of 2

To find the longest wavelength that allows for a photoelectron to be emitted we use the condition

$$
W=frac{hc}{lambda_0}
$$

so we have that the threshold wavelength is given as

$$
lambda_0=frac{hc}{W}=frac{1240times 10^{-9}}{4.5}
$$

Finally, we have that

$$
boxed{lambda_0=276times 10^{-9}textrm{ m}}
$$

Result

2 of 2

$$
lambda_0=276times 10^{-9}textrm{ m}
$$

Exercise 10

Step 1

1 of 2

The reason for this is the fact that frequency is proportional to the photon’s energy and sufficient energy is required to have the excitation which leads to ejection of electrons. Therefore, the low frequency light will not be able eject electrons regardless of its intensity.

Result

2 of 2

The frequency is proportional to the energy and therefore the low-frequency light cannot eject electrons.

Exercise 11

Step 1

1 of 2

Since the frequency of the peak intensity is proportional to the temperature we it will increase with the temperature as

$$
f_{peak}propto T
$$

and the energy is proportional to the temperature

$$
Epropto T^4
$$

So both will increase with the temperature.

Result

2 of 2

Peak frequency and peak energy will rise with the temperature.

Exercise 12

Step 1

1 of 2

By definition this is a result of a photoelectric effect since in a Compton effect another photon has to be emitted, though with a lower frequency than the incident one.

Result

2 of 2

The photoelectric effect.

Exercise 13

Step 1

1 of 2

The photoelectric effect is absorption of a photon by a metal which leads to the emission of electron from it. The Compton effect is an inelastic collision of a photon with an electron where the resulting photon has lower frequency and the electron has higher kinetic energy.

Result

2 of 2

Photoelectric effect- photon is absorbed and an electron is ejected.

Compton effect-photon is scattered by an electron and it exits the collision with a lower frequency.

Exercise 14

Step 1

1 of 2

in order to solve this problem we are going to use the equation of the photoelectric effect which says that

$$
E_{ph}=K+W
$$

So the work function is

$$
W=E_{ph}-K=frac{hc}{lambda}-etimes V_{stop}
$$

$$
W=frac{1.24times 10^{-6}}{0.532times 10^{-6}}-1times 1.44
$$

Finally, we have that

$$
boxed{W=0.89textrm{ eV}}
$$

Result

2 of 2

$$
W=0.89textrm{ eV}
$$

Exercise 15

Step 1

1 of 2

In order to solve this problem we are going to use the definition of the photon’s energy

$$
E=frac{hc}{lambda}=frac{1.24times 10^{-6}}{0.65times 10^{-6}}
$$

Finally, we have that

$$
boxed{E=1.91textrm{ eV}}
$$

Result

2 of 2

$$
E=1.91textrm{ eV}
$$

Exercise 16

Step 1

1 of 2

In order to solve this problem we are going to use the definition of the photon’s energy

$$
E=frac{hc}{lambda}=frac{1.24times 10^{-6}}{0.00002times 10^{-6}}
$$

Finally, we have that

$$
boxed{E=62times 10^3textrm{ eV}}
$$

Result

2 of 2

$$
E=62times 10^3textrm{ eV}
$$

Exercise 17

Step 1

1 of 2

The scattered X ray has a less energy than the initial X ray photon therefore the wavelength of the scattered photon is longer.

Result

2 of 2

longer

Exercise 18

Step 1

1 of 2

To answer these questions we can retrospect the physics of a classical elastic collision which can be used as a good approximation of the Compton effect. We have that the “velocity” of the lighter particle is

$$
v_1=frac{m_1-m_2}{m_1+m_2}u_1
$$

if we take that the initial velocity of the heavier body is zero. So we see that if

$$
m_2gg m_1
$$

the velocity of the lighter object is approximately the same. So we can conclude that the heavier second object is less kinetic energy will have after the collision so the answer to both questions is no i.e. proton will take less energy from the photon and photon will lose less energy when colliding with the proton.

Result

2 of 2

The answer to both questions is no.

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