Physics 6C - University of California, Santa Barbaraclas.sa.ucsb.edu/staff/resource folder/physics6c/Physics... · 2010-06-01 · An easy way to do this one is to find the frequency

Physics 6C

Final Practice Solutions

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Use the following information for problems 1 and 2.A beam of white light with frequency between 4.00 x 1014 Hz and 7.90 x 1014 Hz is incident on a

sodium surface, which has a work function of 2.28 eV.

1) What is the range of frequencies in this beam of light for which electrons are ejected from the sodium surface?

An easy way to do this one is to find the frequency of a photon that has an energy of 2.28 eV.



Hz105.5f

)f()seV1014.4(eV28.2

fhE

14

15

photon

Now we just have to realize that any photon with a higher frequency will have more energy, and will also eject electrons. So the answer is (d).

Use the following information for problems 1 and 2.A beam of white light with frequency between 4.00 x 1014 Hz and 7.90 x 1014 Hz is incident on a

sodium surface, which has a work function of 2.28 eV.

2) What is the maximum kinetic energy of the photoelectrons that are ejected from this surface?

We use the most energetic photon available – the one with the highest frequency. Its energy will be higher than the work function, and the difference will be the leftover KE.



eV99.0KE

eV28.2)Hz109.7()seV1014.4(KE

fhKE

max

1415

max

max

Convert to Joules:

J106.1)106.1()eV99.0(KE 19

eVJ19

max

Answer (b)

3) When light of a particular wavelength and intensity shines on the surface of a given metal, no electrons are ejected. Which of the following changes will most likely cause the emission of photoelectrons?

a) Decreasing the frequency of the lightb) Increasing the intensity of the lightc) Decreasing the wavelength of the lightd) Increasing the wavelength of the light

In order for electrons to be ejected, the photons have to have more energy. To do this we need to either increase the frequency or decrease the wavelength. Changing the intensity will not change the energy of individual photons.

Answer (c).



4) What is the De Broglie wavelength for a Helium atom (atomic mass=4.002603 u) traveling at a speed of 103 m/s? Use the conversion factor 1 u = 1.66054 x 10-27 kg.

This one is just a formula:



m100.1

)10()1066054.1()u002603.4(

sJ1063.6

vm

h

10

sm3

u

kg27

34

Answer (b). You may have to round up carefully on this one.

5) The position of an electron is measured to within +/- 10-6 m. What is the minimum uncertainty in the electron’s velocity? Use electron mass = 9.11x10-31 kg.

We use the uncertainty principle for this one:



s

mkg28

x

x

6

x

05.1p

2

h)p()m10(

2

hpx

At this point recall that momentum is mass x velocity, and divide by the mass of the electron:

sm

x

s

mkg28

x

116v

05.1vm

Answer (c).

6) You and a friend travel through space in identical spaceships. Your friend informs you that he has made some length measurements and that his ship is 150m long and yours is 120m long. What is the relative speed of the two ships?

Use the length contraction formula:



c6.0v

36.0

164.0

18.0

1)m150(m120

1LL

2

2

2

2

2

2

2

2

2

2

c

v

c

v

c

v

c

v

c

v0

Answer (c).

7) A laser is used to move the electron in a hydrogen atom from the n=2 to the n=4 state. What is the wavelength of the light used to excite this electron?

The energy of the absorbed photon is the difference in the energies of the excited states.



eV55.22

eV6.13

4

eV6.13EE

n

eV6.13E

2224

2n

Now find the wavelength of a photon with this energy:

nm490m1087.4

)103()sev1014.4(eV55.2

chE

7

sm815

photon

Answer (b).

8) Assume the excited hydrogen atoms in problem 7 will emit photons until they fall back to the ground state. What is the minimum wavelength of the emitted photons?

Since we want the minimum wavelength, we want the photon with the maximum energy. That will be the one that drop all the way from n=4 to n=1.



eV75.124

eV6.13eV6.13EE

241

Now find the wavelength of a photon with this energy:

nm97m107.9

)103()sev1014.4(eV75.12

chE

8

sm815

photon

Answer (b).

9) The energy required to remove the outermost electron from sodium (Z=11) is 5.1 eV. The energy required to remove the outermost electron from neon (Z=10) is:

a) greater than 5.1 eVb) less than 5.1 eVc) equal to 5.1 eV

d) this problem almost makes me forget about tacos.

Neon has a filled electron shell, thus has a much higher ionization energy than sodium, which has a single electron in the 3s shell.

Answer (a).

As a general rule (with a few exceptions) ionization energies will increase as you go to the right along any row of the periodic table, and will decrease as you go down any column.



10) A 239Pu (plutonium) nucleus typically decays by emitting an α-particle with an energy of about 5.1 MeV. What is the resulting daughter nucleus?

Alpha-decay ejects a helium nucleus from the nucleus of the large atom, leaving it with 2 fewer protons, and 4 fewer nucleons overall. So we need to use the periodic table and find the element that has 2 fewer protons than plutonium.

This is uranium. The number of nucleons should be 4 fewer, so we get 235U.

Answer (a).



11) Nucleus X has 10 times as many nucleons as nucleus Y. Compared to Y, X hasa) a larger radius and about the same densityb) a larger radius and a smaller densityc) a smaller radius and about the same densityd) a larger radius and a larger density

We have a formula for nuclear radius:



)A()m102.1(R 3115

So nucleus X should have a larger radius since it has a larger value of A.

However, nuclear density is about the same for all nucleons (we did this in class).

Answer (a).

12) Originally the decay rate of a sample was 16 decays per second. Now it has a rate of 2 decays per second. The half-life is 5 years. How many years have passed?

If you notice that the numbers come out perfectly, then you can do this one without much calculating. The decay rate has dropped from 16 to 2. This is a factor of 8. Since 23=8, we know the time should be 3 half-lives, or 15 years.

Answer (d).



13) An ancient bone has been recovered from an archeological dig. It is determined that the ratio of the isotope carbon-14 to total carbon is only 20% of the ratio currently present in the atmosphere. The half-life of carbon-14 is 5730 years. How old is the bone?

We will need to use our exponential decay formula for this one.

The decay constant comes from the half-life:



t0 eN)t(N

yr141021.1

T

)2ln(

21

yr300,13t

t)1021.1()2.0ln(

e2.0

eNN%)20(

yr14

t

t

00

Answer (c).

14) 167N is an unstable isotope of nitrogen with a half-life of about 7 seconds. When it undergoes

- decay an electron, an antineutrino, and one other particle are emitted. What is the other particle?

The other particle will be the daughter nucleus, which should have one more proton than the original nitrogen atom. From looking at the periodic table, we find that it must be oxygen.

Answer (b).



15) Which of the following is a valid set of quantum numbers to describe the outermost valence electron in the Fluorine atom?

a) n = 2, l = 1, ml = -1, ms = +1/2

b) n = 2, l = 0, ml = +1, ms = -1/2

c) n = 1, l = 0, ml = -1, ms = +1/2

d) n = 2, l = 2, ml = +1, ms = +1/2

(a) is the only one of these options that is valid.

(b) and (c) violate the rule that ml cannot be larger than l.

(d) violates the rule that l cannot be larger than n-1.



16) An oil film floats on water. The refractive index of the oil is 1.4 and the index for water is 1.33. The film appears blue because it reflects light with wavelength 400 nm very well. Calculate the minimum thickness of the oil film.



Air n=1

Oil n=1.40

Water n=1.33

12Ray 1 reflects from a higher index, but ray 2 reflects

from a lower index, so there is a relative phase shift.

Our formulas are:

veconstructit2n

)m(

edestructivt2n

m

021

0

We use the constructive formula, with m=0 to get the thinnest possible film:

nm71t

t24.1

nm400)0(

min

21

Answer (a).

17) A double-slit experiment is performed with light of wavelength of 500 nm, and the first dark fringe appears at a distance of 4cm from the central bright maximum. If this experiment is now performed with the same setup, but underwater, how far from center will the first dark fringe be? Assume the index of refraction of water is 1.33.

This is a double-slit experiment, so we can use the approximate formula for the position of the dark fringes. Use m=0 to get the first dark fringe from center.



d

)m(Ry 2

1

m

Before we plug in the numbers, realize that the only difference between the experiments will be the wavelength of the light. Underwater, the wavelength will be shortened. So we can set up a ratio to solve for the distance.

cm333.1

cm4y

n

yy

n

1

dR

dnR

y

y

dn

)0(Ry

d

)0(Ry

water

airwater

021

021

air

water

021

water

021

air

Answer (b).

18) Two telescopes are to be placed in orbit. One telescope is designed to be used with visible light of wavelength 500nm, while the second telescope is designed to be used with infrared light of wavelength 2000nm. In order that both telescopes have the same resolution, the mirror of the infrared telescope must have

a) 1/4 the diameter of the visible-light telescope’s mirrorb) 2 times the diameter of the visible-light telescope’s mirrorc) 4 times the diameter of the visible-light telescope’s mirrord) 16 times the diameter of the visible-light telescope’s mirror

Resolution depends on the wavelength of light and the diameter of the of the light collector:



D22.1res

Since the infrared telescope is using light with 4 times the wavelength, it requires a mirror with 4 times the diameter to get the same resolution.

Answer (c).

19) The sun produces energy by nuclear fusion reactions, in which matter is converted to energy. The rate of energy production is 3.8 x 1026 Watts. How many kilograms of mass does the sun convert to energy each second?

Recall that a Watt is a Joule/Second.

Convert energy to mass using Einstein’s formula:



kg102.4m

)103()m(J108.3

cmE

9

2

sm826

2

Answer (a).

20) Calculate the mass defect for the decay of 84Be into two alpha particles.The mass of 84Be is 8.005305 u, and the mass of an alpha particle is 4.00148 u.Use the conversion factor 1 u = 931.5 MeV

We need to find the difference between the mass of the original Be nucleus (don’t forget to account for the electrons!), and the mass of the 2 alpha particles that result from the decay:



MeV14.0)5.931()u000149.0(m

u000149.0)u00148.4(2)u000549.0(4u005305.8m

uMeV

nucleusBeryllium

Answer (c).

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Physics 6C - University of California, Santa Barbaraclas.sa.ucsb.edu/staff/resource folder/physics6c/Physics... · 2010-06-01 · An easy way to do this one is to find the frequency