Home Page Textbooks Physics: Principles and Problems Page 745: Standardized Test Practice

Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

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Page 745: Standardized Test Practice

Exercise 1

Step 1

1 of 3

The quantized energy levels of an atom are integer multiples of the ground or minimum energy level.

Step 2

2 of 3

Therefore any non-integer multiple of the basic quantum of energy is impossible state. We see that in our case that is $frac{3}{4}hf$ or the choice A.

Result

3 of 3

The choice A is not possible.

Exercise 2

Step 1

1 of 2

D) It is the minimum frequency of incident radiation needed to cause the ejection of electrons from an atom.

Result

2 of 2

Exercise 3

Step 1

1 of 3

In this problem, we have a photon of a given frequency for which we should find the corresponding energy.

Step 2

2 of 3

We can do so by employing the Planck equation which relates the frequency with the energy as follows

$$
E_{ph}=hf=6.63times 10^{-34}times 1.14times 10^{15}
$$

So the answer is

$$
boxed{textrm{B) }E_{ph}=7.55times 10^{-19} textrm{ J}}
$$

Result

3 of 3

$$
textrm{B) }E_{ph}=7.55times 10^{-19} textrm{ J}
$$

Exercise 4

Step 1

1 of 3

In this problem we have a radiation of a given energy that is hitting the photocell with a given work function.

Step 2

2 of 3

To solve it, we are going to employ Einstein’s photoelectric effect equation which tells us that the kinetic energy of an ejected electron is given as

$$
K=E_{ph}-W=5.17-2.31
$$

So the answer is

$$
boxed{textrm{C) }K=2.86textrm{ eV}}
$$

Result

3 of 3

$$
textrm{C) }K=2.86textrm{ eV}
$$

Exercise 5

Step 1

1 of 5

In this problem we have an electron which is being accelerated across the given potential difference. Our task is to find its de Broglie wavelength.

Step 2

2 of 5

To do so, we should first find the electrons speed which can be found from the conservation of energy law which tells us that at the end of the acceleration it has the hold that

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

Step 3

3 of 5

From the previous expression one can easily express the speed as follows

$$
v=sqrt{frac{2qV}{m}}
$$

Step 4

4 of 5

Now, we can use this and de Broglie equation to find the de Broglie wavelength of the electron. Namely

$$
lambda=frac{h}{mv}=frac{h}{sqrt{2qmV}}=frac{6.63times 10^{-34}}{sqrt{2times 1.6times 10^{-19} times 9.11times 10^{-31}times 95 }}
$$

Finally, we have that

$$
boxed{textrm{B) }lambda=1.26times 10^{-10}textrm{ m}}
$$

Result

5 of 5

$$
textrm{B) }lambda=1.26times 10^{-10}textrm{ m}
$$

Exercise 6

Step 1

1 of 4

In this problem we have an electron of a given speed for which we should find de Broglie wavelength.

Step 2

2 of 4

We can do so by simply applying the de Broglie equation which tells us that

$$
lambda=frac{h}{mv}
$$

Step 3

3 of 4

Now, we can plug in the known values in the de Broglie equation to obtain that

$$
lambda=frac{6.63times 10^{-34}}{9.11times 10^{-31}times 391times 10^3}
$$

Finally, we have that

$$
boxed{lambda=1.86times 10^{-9}textrm{ m}}
$$

so the correct answer is under choice D.

Result

4 of 4

$$
textrm{D) }lambda=1.86times 10^{-9}textrm{ m}
$$

Exercise 7

Step 1

1 of 2

According to Einstein’s equation of photoelectric effect we have that the work function is the energy that has to be paid to free the electron most weakly bound to the metal. Therefore the answer is D.

Result

2 of 2

The answer is D.

Exercise 8

Step 1

1 of 4

In this problem we have an object with a given de Brogile wavelength and the speed and we are trying to find its mass.

Step 2

2 of 4

To do so, we are going to use the de Broglie equation from which we can express the mass.

Step 3

3 of 4

So we start with the aforementioned expression from which we will obtain the mass

$$
lambda=frac{h}{mv}
$$

The mass is then given as

$$
m=frac{h}{vlambda}=frac{6.63times 10^{-34}}{2.3times 10^{-34}times 45}
$$

$$
boxed{m=0.064 textrm{ kg} }
$$

Result

4 of 4

$$
m=0.064 textrm{ kg}
$$

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