Home Page Textbooks Physics: Principles and Problems Page 757: Practice Problems

Physics: Principles and Problems

9th Edition

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

ISBN: 9780078458132

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Page 757: Practice Problems

Exercise 1

Step 1

1 of 6

In this problem we should calculate the energy of the second, third and fourth energy level in hydrogen atom.

Step 2

2 of 6

To do so, we are going to employ a simple yet powerful formula of the energy levels in the hydrogen atom

$$
E_n=-frac{13.6}{n^2}
$$

Step 3

3 of 6

Now we can use this and in the case when $n=2$ the above equation simply becomes

$$
E_2=-frac{13.6}{4}=boxed{-3.4textrm{ eV}}
$$

Step 4

4 of 6

Furthermore, we obtain that in the case when $n=3$ the above equation simply becomes

$$
E_3=-frac{13.6}{9}=boxed{-1.51textrm{ eV}}
$$

Step 5

5 of 6

Finally, when $n=4$ one obtains that the energy level is given as

$$
E_4=-frac{13.6}{16}=boxed{-0.85textrm{ eV}}
$$

Result

6 of 6

$$
E_2=-3.4textrm{ eV}
$$

$$
E_3=-1.51textrm{ eV}
$$

$$
E_4=-0.85textrm{ eV}
$$

Exercise 2

Step 1

1 of 4

In this problem we have to find the energy difference between $E_3$ and $E_2$ in the hydrogen atom.

Step 2

2 of 4

We can do so by applying the fact that the each energy level in the hydrogen atom is quantized as

$$
E_n=-frac{13.6}{n^2}textrm{ eV}
$$

Step 3

3 of 4

Now, in the case of the the levels given in the problem we have that

$$
Delta E=E_3-E_2=-13.6times(frac{1}{9}-frac{1}{4})=frac{5}{36}times 13.6
$$

Finally, we have that

$$
boxed{Delta E =1.89textrm{ eV}}
$$

Result

4 of 4

$$
Delta E =1.89textrm{ eV}
$$

Exercise 3

Step 1

1 of 4

In this problem we have to find the energy difference between $E_4$ and $E_2$ in the hydrogen atom.

Step 2

2 of 4

We can do so by applying the fact that the each energy level in the hydrogen atom is quantized as

$$
E_n=-frac{13.6}{n^2}textrm{ eV}
$$

Step 3

3 of 4

Now, in the case of the the levels given in the problem we have that

$$
Delta E=E_4-E_2=-13.6times(frac{1}{16}-frac{1}{4})=frac{3}{16}times 13.6
$$

Finally, we have that

$$
boxed{Delta E =2.55textrm{ eV}}
$$

Result

4 of 4

$$
Delta E =2.55textrm{ eV}
$$

Exercise 4

Step 1

1 of 7

In this problem we should find the radii of the second, third and fourth energetic orbits in the hydrogen atom.

Step 2

2 of 7

To do so, we will use the formula given in the problem itself which gives the dependence of the radius on the energetic level

$$
r_n=frac{h^2n^2}{4pi^2Kmq^2}
$$

and the fact that

$$
r_1=5.3times 10^{-11}textrm{ m}
$$

Step 3

3 of 7

Namely, one easily observes that radius of any discrete energy level in the hydrogen atom can be expressed as

$$
r_n=n^2r_1
$$

Step 4

4 of 7

So in the case of $n=2,3,4$ we are going to have that

$$
r_2=4times r_1=4times 5.3times 10^{-11}=boxed{21.2times 10^{-11}textrm{ m}}
$$

Step 5

5 of 7

$$
r_3=9times r_1=9times 5.3times 10^{-11}=boxed{47.7times 10^{-11}textrm{ m}}
$$

Step 6

6 of 7

$$
r_4=16times r_1=16times 5.3times 10^{-11}=boxed{84.8times 10^{-11}textrm{ m}}
$$

Result

7 of 7

$$
r_2=21.2times 10^{-11}textrm{ m}
$$

$$
r_3=47.7times 10^{-11}textrm{ m}
$$

$$
r_4=84.8times 10^{-11}textrm{ m}
$$

Exercise 5

Solution 1

Solution 2

Step 1

1 of 4

In this problem we are asked a hypothetical question of how far away would the electron be of a ball of 7.5cm diameter was used as the nucleus.

Step 2

2 of 4

To solve this problem we are going to use the ratio of the values of different quantities. Let’s $d_0$ is the diameter of the nucleus, $a_0$ is the radius of the hydrogen atom in its ground state we have that

$$
frac{d_0}{a_0}=frac{d_{ball}}{X}
$$

Step 3

3 of 4

From the above expression one can easily express the hypothetical distance to the electron as

$$
X=frac{d_{ball}}{d_0}a_0=frac{7.5times 10^{-2}}{2.5times 10^{-15}}times 5times 10^{-11}
$$

Finally, we have that

$$
boxed{X=1.5times 10^3textrm{m}}
$$

Result

4 of 4

$$
X=1.5times 10^3textrm{m}
$$

Step 1

1 of 6

$dfrac{2.5*10^{-15}}{2}= 1.25times 10^{-15}$ m

to get the radius we divide the diameter by 2

Step 2

2 of 6

$dfrac{7.5}{2}=3.75$ cm

to get the radius we divide the diameter by 2
but the result is in cm we need to convert it to meter
so we multipy by $10^{-2}$

Step 3

3 of 6

$3.75 times 10^{-2}$ m

radius of ball

Step 4

4 of 6

$$
dfrac{1.25times 10^{-15}}{5times 10{-11}}= dfrac{3.275times 10^{-2}}{X}
$$

general relativity formula

Step 5

5 of 6

$X = dfrac{5times 10^{-11}times 3.75 times 10^{-2}}{1.25 times 10^{-15}}=1500$ m

Result

6 of 6

distance = 1500 m

Exercise 6

Step 1

1 of 8

In this problem we are asked to find the wavelengths of photons emitted during $E_{32}$ and $E_{42}$ transitions. However we first have to find the corresponding energies.

Step 2

2 of 8

We can do so by applying the fact that the each energy level in the hydrogen atom is quantized as

$$
E_n=-frac{13.6}{n^2}textrm{ eV}
$$

Step 3

3 of 8

a) Let’s do it first for the case $E_{32}$ by simply following the equation above

$$
E_{32}=E_3-E_2=-13.6times(frac{1}{9}-frac{1}{4})=frac{5}{36}times 13.6
$$

Finally, we have that

$$
E_{32} =1.89textrm{ eV}
$$

Step 4

4 of 8

Now, the wavelength of the photon can be found by the relation

$$
E_{ph}=frac{hc}{lambda}rightarrowlambda=frac{hc}{E_{ph}}
$$

where $hc=1240times 10^{-9}$eV$cdot$m

Step 5

5 of 8

By plugging in the values. the wavelength in the first case becomes then

$$
lambda=frac{1240times 10^{-9}}{1.89}=boxed{656times 10^{-9} textrm{ m}}
$$

Step 6

6 of 8

b) Now, in the case of $E_{42}$

$$
E_{42}=E_4-E_2=-13.6times(frac{1}{16}-frac{1}{4})=frac{3}{16}times 13.6
$$

Finally, we have that

$$
E_{42} =2.55textrm{ eV}
$$

Step 7

7 of 8

By plugging in the values. the wavelength in the first case becomes then

$$
lambda=frac{1240times 10^{-9}}{2.55}=boxed{486times 10^{-9} textrm{ m}}
$$

Result

8 of 8

$$
textrm{a) }lambda=656times 10^{-9} textrm{ m}
$$

$$
textrm{b) }lambda=486times 10^{-9} textrm{ m}
$$

Exercise 7

Step 1

1 of 5

In this problem we have a transition between two energy levels of given energies in the mercury atom and a photon emitted as a result. We are aiming to find the energy and the wavelength of the emitted photon.

Step 2

2 of 5

a) The energy of the emitted photon can be simply found from the difference between the two energy levels

$$
E_{ph}=E_m-E_n=8.82-6.67=boxed{2.15textrm{ eV}}
$$

Step 3

3 of 5

b) Now, the wavelength of the photon can be found directly from the Planck relation

$$
lambda=frac{hc}{Delta E}
$$

where $hc=1240times 10^{-9}$eV$cdot$m.

Step 4

4 of 5

We can plug in the given and obtained values to get that the wavelength of the emitted photon is

$$
lambda=frac{1240times 10^{-9}}{2.15}=boxed{576times 10^{-9}textrm{ m}}
$$

Result

5 of 5

$$
textrm{a) }E_{ph}=2.15textrm{ eV}
$$

$$
textrm{b) }lambda=576times 10^{-9}textrm{ m}
$$

Exercise 8

Step 1

1 of 5

In this problem we are dealing with a helium atom which undergoes a transition to its ground state while emitting a photon of a given wavelength. We are trying to find the energy of the excited state of the atom.

Step 2

2 of 5

We can solve this problem by firstly calculating the energy of the emitted photon and then adding it to the energy of the ground state. The first part we do by using the formula

$$
E_{ph}=frac{hc}{lambda}
$$

where $hc=1240times 10^{-9}$eV$cdot$m.

Step 3

3 of 5

Now, we can substitute the given values into the relation given above to obtain that

$$
E_{ph}=frac{1240times 10^{-9}}{304times 10^{-9}}=4.08textrm{ eV}
$$

Step 4

4 of 5

Now, the energy of the original, excited state is given directly from

$$
E_{m}=E_1+E_{ph}=-54.4+4.08=boxed{-50.3textrm{ eV}}
$$

Result

5 of 5

$$
E_{m}=-50.3textrm{ eV}
$$

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Page 757: Practice Problems

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