PHY 250 General Physics: Exam Questions and Answers Latest Update
2023/2024 | Graded A+ | Straighterline
HW1. A slow freight train chugs along a straight track. The distance it has traveled
after x hours is given by a function g(x). An engineer is walking along the top of the
box cars at the rate of 6 km/hr in the same direction as the train is moving. The
speed of the man (in km/hr) relative to the ground is: - ANSWER g'(x) + 6
v(Eng/Ground) = v(Eng/Train) + v(Train/Ground)
HW2. Explain why accelerating charges generate light but charges that are
stationary or moving at a constant velocity do not. - ANSWER NOTICE Light is a
wave: an oscillation in time and space of the E and B fields.
A stationary charge produces a time dependent E-field.
A constant velocity charge constitutes a current and produces a time-independent B-
field.
Therefore it is necessary for a charge to accelerate to produce light.
Note: One can do work on a charge by "dragging" it through an E-field at
CONSTANT velocity. Thus, in general, doing work on a charge is NOT enough to
guarantee production of light!
HW2. It is the thermal motion of charged particles at the sun's surface that produces
the electromagnetic radiation emitted by the sun (use c=3.0 E8 m/s). To generate a
blue light at 400nm, at what frequency would a charged particle have to be vibrating
back and forth? - ANSWER f = c/lambda = 7.5e14 Hz
HW2. It is the thermal motion of charged particles at the sun's surface that produces
the electromagnetic radiation emitted by the sun (use c=3.0 E8 m/s). To generate a
blue light at 400nm, a charged particle have to be vibrating back and forth at a very
high frequency.
Even in the most advanced circuits, we cannot oscillate electrons back and forth at
that rate through wires. But we can oscillate charges back and forth quickly enough
to broadcast TV using radiowave signals. At what frequency does that electronics at
the TV station need to have the charges oscillate back and forth on a TV broadcast
antenna to transmit a typical TV signal (say a radiowave transmission signal with a
wavelength of 3 meter)? - ANSWER f = c/lambda = 3.0e8/3 = 10e8 Hz = 100 MHz
Learning Goal (May 08 Lecture): Write down the mathematical description of a
classical electromagnetic wave, and relate the terms in the expression to the
velocity, wavelength,
and frequency of the wave. - ANSWER
Learning Goal (May 08 Lecture): Describe the energy in a classical EM wave in
terms of the amplitude of the wave, and describe quantitatively what happens when
an EM wave is absorbed by a material if the wavelength is long enough and intensity
high enough that it behaves classically. - ANSWER
,HW2. When we were discussing the classical wave-view of electromagnetic light, we
considered the following scenario where 3 different beams of laser light (single-
frequency light) were hitting 3 barrels filled with water. The drawing showed the
frequency and amplitude of the electromagnetic wave in each case (the amplitude
for #1 and #2 are the same). The clicker question compared how fast the barrels will
heat. T or F: The total amount of power hitting barrel #1 is less than the total amount
hitting barrel #2. - ANSWER False
The power is proportional to the maximum E-field multiplied by the effective surface
area in contact with the light. Since this area is the same for both barrels, as well as
Emax, the power must be the same.
(May 10 Lecture): T or F. The amount of energy in each photon hitting barrel #1 is
the same as the amount in each photon hitting barrel #3. - ANSWER True
The two waves have the same wavelength and hence the same frequency (f*lamba
= c). Thus E(photon) = hf is the same in both cases.
(May 10 Lecture): T or F. The number of photons hitting barrel #1 per second is the
same as the number of photons hitting barrel #2 per second. - ANSWER False
The total number of photons hitting the barrel per unit time multiplied by the energy
per photon is actually the total power delivered. These waves have the same power
but different frequency and hence different energy per photon. Therefore, the
number of photons hitting the barrels must be different.
(May 10 Lecture): T or F. Barrel #1 and #2 heat up at the same rate. - ANSWER
True
The rate of heating must be proportional to the power, since this is the energy
delivered per unit time. These waves have the same power.
(May 10 Lecture): From the picture you can see that, (wavelength of #2) = 3/5
(wavelength of #1). If there are 2,500,000 photons per second hitting barrel #2, how
many photons are hitting barrel #1 per second? - ANSWER Power = (number of
photons(n)*energy of photon(E))/change in time(dt)
Here dt = 1 second
E = hf
We know P1 = P2
n1E1/dt = n2E2/dt
n1 = n2*(hf2/hf1) = n2*(lambda1/lambda2)
n2 = 2500000
n1 = 250000*(5/3)
, HW2. A photoelectric-effect experiment finds a stopping potential of 1.93 V when
light of 200 nm is used to illuminate the cathode. From what metal is the cathode
made? (hint, use table 39.1 in Knight Volume 5) - ANSWER Aluminum
The stopping potential must be the right strength to stop even the most energetic
electrons, which have energy E(photon) - E.
So we need Vstop = W/e = deltaKEmax/e = (Ephoton-E0)/e = (hf-E0)/e
E0 = hf-eVstop = 4.27
HW2. A photoelectric-effect experiment finds a stopping potential of 1.93 V when
light of 200 nm is used to illuminate the cathode. The intensity of the light is doubled.
What is the stopping potential now? - ANSWER The stopping potential depends only
on frequency, work function, and change of electron, NOT on intensity!
Stationary Charges - ANSWER constant E-field, no magnetic (B)-field
(We don't see charges glow in the dark)
Charges moving at a constant velocity - ANSWER Constant current through wire
creates a B-field
but B-field is constant. (We don't SEE DC.)
Accelerating charges - ANSWER changing E-field and changing B-field
(EM radiation...both E and B are oscillating)
We talked briefly about Maxwell equations
HW5. The process where a photon hits an atom that is already in a higher energy
level and this causes the atom to spit out a photon that is identical to the one that hit
the atom resulting in two identical photons - ANSWER Stimulated emission
The fact that a passing photon stimulates the atom to emit another photon explains
the name of this process
HW5. The process by which the light is absorbed and the energy causes the atomic
electron to go to a higher energy level - ANSWER Absorption
If an electron jumps to a higher energy state, it must have absorbed a photon
HW5. The process in which the electron naturally jumps down from a higher energy
state to a lower state and spits out a photon corresponding to the energy difference
as it does so - ANSWER Spontaneous Emission
If the electron jumps down on its own, without being influenced by a passing photon,
it is called spontaneous emission
HW5. A laser is created by producing a lot of photons created by stimulated
emission. What characteristics make light produced by stimulated emission so
special compared to light produced through spontaneous emission? - ANSWER -
Photons' electromagnetic waves are oscillating exactly in phase.
2023/2024 | Graded A+ | Straighterline
HW1. A slow freight train chugs along a straight track. The distance it has traveled
after x hours is given by a function g(x). An engineer is walking along the top of the
box cars at the rate of 6 km/hr in the same direction as the train is moving. The
speed of the man (in km/hr) relative to the ground is: - ANSWER g'(x) + 6
v(Eng/Ground) = v(Eng/Train) + v(Train/Ground)
HW2. Explain why accelerating charges generate light but charges that are
stationary or moving at a constant velocity do not. - ANSWER NOTICE Light is a
wave: an oscillation in time and space of the E and B fields.
A stationary charge produces a time dependent E-field.
A constant velocity charge constitutes a current and produces a time-independent B-
field.
Therefore it is necessary for a charge to accelerate to produce light.
Note: One can do work on a charge by "dragging" it through an E-field at
CONSTANT velocity. Thus, in general, doing work on a charge is NOT enough to
guarantee production of light!
HW2. It is the thermal motion of charged particles at the sun's surface that produces
the electromagnetic radiation emitted by the sun (use c=3.0 E8 m/s). To generate a
blue light at 400nm, at what frequency would a charged particle have to be vibrating
back and forth? - ANSWER f = c/lambda = 7.5e14 Hz
HW2. It is the thermal motion of charged particles at the sun's surface that produces
the electromagnetic radiation emitted by the sun (use c=3.0 E8 m/s). To generate a
blue light at 400nm, a charged particle have to be vibrating back and forth at a very
high frequency.
Even in the most advanced circuits, we cannot oscillate electrons back and forth at
that rate through wires. But we can oscillate charges back and forth quickly enough
to broadcast TV using radiowave signals. At what frequency does that electronics at
the TV station need to have the charges oscillate back and forth on a TV broadcast
antenna to transmit a typical TV signal (say a radiowave transmission signal with a
wavelength of 3 meter)? - ANSWER f = c/lambda = 3.0e8/3 = 10e8 Hz = 100 MHz
Learning Goal (May 08 Lecture): Write down the mathematical description of a
classical electromagnetic wave, and relate the terms in the expression to the
velocity, wavelength,
and frequency of the wave. - ANSWER
Learning Goal (May 08 Lecture): Describe the energy in a classical EM wave in
terms of the amplitude of the wave, and describe quantitatively what happens when
an EM wave is absorbed by a material if the wavelength is long enough and intensity
high enough that it behaves classically. - ANSWER
,HW2. When we were discussing the classical wave-view of electromagnetic light, we
considered the following scenario where 3 different beams of laser light (single-
frequency light) were hitting 3 barrels filled with water. The drawing showed the
frequency and amplitude of the electromagnetic wave in each case (the amplitude
for #1 and #2 are the same). The clicker question compared how fast the barrels will
heat. T or F: The total amount of power hitting barrel #1 is less than the total amount
hitting barrel #2. - ANSWER False
The power is proportional to the maximum E-field multiplied by the effective surface
area in contact with the light. Since this area is the same for both barrels, as well as
Emax, the power must be the same.
(May 10 Lecture): T or F. The amount of energy in each photon hitting barrel #1 is
the same as the amount in each photon hitting barrel #3. - ANSWER True
The two waves have the same wavelength and hence the same frequency (f*lamba
= c). Thus E(photon) = hf is the same in both cases.
(May 10 Lecture): T or F. The number of photons hitting barrel #1 per second is the
same as the number of photons hitting barrel #2 per second. - ANSWER False
The total number of photons hitting the barrel per unit time multiplied by the energy
per photon is actually the total power delivered. These waves have the same power
but different frequency and hence different energy per photon. Therefore, the
number of photons hitting the barrels must be different.
(May 10 Lecture): T or F. Barrel #1 and #2 heat up at the same rate. - ANSWER
True
The rate of heating must be proportional to the power, since this is the energy
delivered per unit time. These waves have the same power.
(May 10 Lecture): From the picture you can see that, (wavelength of #2) = 3/5
(wavelength of #1). If there are 2,500,000 photons per second hitting barrel #2, how
many photons are hitting barrel #1 per second? - ANSWER Power = (number of
photons(n)*energy of photon(E))/change in time(dt)
Here dt = 1 second
E = hf
We know P1 = P2
n1E1/dt = n2E2/dt
n1 = n2*(hf2/hf1) = n2*(lambda1/lambda2)
n2 = 2500000
n1 = 250000*(5/3)
, HW2. A photoelectric-effect experiment finds a stopping potential of 1.93 V when
light of 200 nm is used to illuminate the cathode. From what metal is the cathode
made? (hint, use table 39.1 in Knight Volume 5) - ANSWER Aluminum
The stopping potential must be the right strength to stop even the most energetic
electrons, which have energy E(photon) - E.
So we need Vstop = W/e = deltaKEmax/e = (Ephoton-E0)/e = (hf-E0)/e
E0 = hf-eVstop = 4.27
HW2. A photoelectric-effect experiment finds a stopping potential of 1.93 V when
light of 200 nm is used to illuminate the cathode. The intensity of the light is doubled.
What is the stopping potential now? - ANSWER The stopping potential depends only
on frequency, work function, and change of electron, NOT on intensity!
Stationary Charges - ANSWER constant E-field, no magnetic (B)-field
(We don't see charges glow in the dark)
Charges moving at a constant velocity - ANSWER Constant current through wire
creates a B-field
but B-field is constant. (We don't SEE DC.)
Accelerating charges - ANSWER changing E-field and changing B-field
(EM radiation...both E and B are oscillating)
We talked briefly about Maxwell equations
HW5. The process where a photon hits an atom that is already in a higher energy
level and this causes the atom to spit out a photon that is identical to the one that hit
the atom resulting in two identical photons - ANSWER Stimulated emission
The fact that a passing photon stimulates the atom to emit another photon explains
the name of this process
HW5. The process by which the light is absorbed and the energy causes the atomic
electron to go to a higher energy level - ANSWER Absorption
If an electron jumps to a higher energy state, it must have absorbed a photon
HW5. The process in which the electron naturally jumps down from a higher energy
state to a lower state and spits out a photon corresponding to the energy difference
as it does so - ANSWER Spontaneous Emission
If the electron jumps down on its own, without being influenced by a passing photon,
it is called spontaneous emission
HW5. A laser is created by producing a lot of photons created by stimulated
emission. What characteristics make light produced by stimulated emission so
special compared to light produced through spontaneous emission? - ANSWER -
Photons' electromagnetic waves are oscillating exactly in phase.
