Quantum and Nuclear Physics

1

A monochromatic light source is incident on a neutral density filter which decreases the light source's intensity by a factor of 3 but does not change its color. What happens to the energy of a photon in the less intense beam?

None of these

The photon energy decreases by a factor of 3.

The photon energy increases by a factor of 3.

The photon energy decreases by a factor of 9.

The photon energy decreases only slightly.

Explanation

The photon energy does not change since the color of light (and therefore the frequency of the light) does not change.

2

Radiocarbon dating is a method that allows scientists to estimate the age of organisms after they have died. A fairly constant amount of radioactive $C^{14}$ remains in the animal while alive, but once it dies the $C^{14}$ degrades over time into $C^{12}$ . By measuring the relative amount of these two in a dead organism compared to a living one, along with knowing that the half-life of $C^{14}$ is years, it can be determined how long ago the organism died.

If a scientist finds that a fossil contains amount of the $C^{14}$ normally found in the atmosphere at that time, approximately how old is this fossil?

Explanation

In this question, we're given a brief description of radiocarbon dating. We're given the amount of $C^{14}$ that has been found in a fossil sample and we're asked to find the approximate age of the fossil.

First, let's briefly go over radiocarbon dating. This method essentially assumes that the amount of radioactive carbon within an organism remains fairly stable at any given time while the organism is alive. Moreover, this amount of radioactive carbon is related to the amount of radioactive carbon in the atmosphere. Once the organism dies, however, it ceases to gain any radioactive carbon; rather, the $C^{14}$ that was present now begins to decay into $C^{12}$ . Thus, by measuring the amount of $C^{14}$ in the organism and comparing it to the amount in the atmosphere, the age at which the organism died can be approximated.

Since we're dealing with the decay of $C^{14}$ , this is a radioactive decay problem. Recall that all radioactive decay reactions follow first-order rate kinetics. What this means is that the rate of decay is only dependent on the amount of radioactive material at any given instant. Hence, we can use the first-order rate equation.

$A_{t}=A_{0}e^{-kt}$

We can further rearrange this expression to isolate the variable for time.

$\frac{A_t}{A_0}=e^{-kt}$

$t=-\frac{ln(\frac{A_t}{A_0})}{k}$

In arriving at this expression, we see that we need to know the rate constant, , in order to solve for . To do this, we can use the equation that relates the half-life to the rate constant for a first-order process.

$t_\frac{1}{2}=\frac{ln(2)}{k}$

Rearranging, we can find the rate-constant.

$k=\frac{ln(2)}{t_\frac{1}{2}}$

$k=\frac{0.693}{5730: yr}=1.21\cdot 10^{-4}: yr^{-1}$

Now that we have the rate constant, we can plug this value into the previous expression to solve for .

$t=-\frac{ln(\frac{0.35}{1})}{1.21\cdot 10^{-4}: yr^{-1}}=8676:years$

3

Radiocarbon dating is a method that allows scientists to estimate the age of organisms after they have died. A fairly constant amount of radioactive $C^{14}$ remains in the animal while alive, but once it dies the $C^{14}$ degrades over time into $C^{12}$ . By measuring the relative amount of these two in a dead organism compared to a living one, along with knowing that the half-life of $C^{14}$ is years, it can be determined how long ago the organism died.

If a scientist finds that a fossil contains amount of the $C^{14}$ normally found in the atmosphere at that time, approximately how old is this fossil?

Explanation

In this question, we're given a brief description of radiocarbon dating. We're given the amount of $C^{14}$ that has been found in a fossil sample and we're asked to find the approximate age of the fossil.

First, let's briefly go over radiocarbon dating. This method essentially assumes that the amount of radioactive carbon within an organism remains fairly stable at any given time while the organism is alive. Moreover, this amount of radioactive carbon is related to the amount of radioactive carbon in the atmosphere. Once the organism dies, however, it ceases to gain any radioactive carbon; rather, the $C^{14}$ that was present now begins to decay into $C^{12}$ . Thus, by measuring the amount of $C^{14}$ in the organism and comparing it to the amount in the atmosphere, the age at which the organism died can be approximated.

Since we're dealing with the decay of $C^{14}$ , this is a radioactive decay problem. Recall that all radioactive decay reactions follow first-order rate kinetics. What this means is that the rate of decay is only dependent on the amount of radioactive material at any given instant. Hence, we can use the first-order rate equation.

$A_{t}=A_{0}e^{-kt}$

We can further rearrange this expression to isolate the variable for time.

$\frac{A_t}{A_0}=e^{-kt}$

$t=-\frac{ln(\frac{A_t}{A_0})}{k}$

In arriving at this expression, we see that we need to know the rate constant, , in order to solve for . To do this, we can use the equation that relates the half-life to the rate constant for a first-order process.

$t_\frac{1}{2}=\frac{ln(2)}{k}$

Rearranging, we can find the rate-constant.

$k=\frac{ln(2)}{t_\frac{1}{2}}$

$k=\frac{0.693}{5730: yr}=1.21\cdot 10^{-4}: yr^{-1}$

Now that we have the rate constant, we can plug this value into the previous expression to solve for .

$t=-\frac{ln(\frac{0.35}{1})}{1.21\cdot 10^{-4}: yr^{-1}}=8676:years$

4

A car with mass is heading toward a wall at a speed $25\frac{m}{s}$ . What is the approximate de Broglie wavelength of the car?

$h = 6.626\times 10^{-34}~\text{J}\cdot\text{s}$

$1.33\times10^{-38}~\text{m}$

$2\times10^{-34}~\text{m}$

$80~\text{m}$

$0.0125~\text{m}$

$532\times10^{-9}~\text{m}$

Explanation

The de Broglie wavelength is given by $\lambda = \frac{h}{m v}$ .

5

A monochromatic light source is incident on a neutral density filter which decreases the light source's intensity by a factor of 3 but does not change its color. What happens to the energy of a photon in the less intense beam?

None of these

The photon energy decreases by a factor of 3.

The photon energy increases by a factor of 3.

The photon energy decreases by a factor of 9.

The photon energy decreases only slightly.

Explanation

The photon energy does not change since the color of light (and therefore the frequency of the light) does not change.

6

A monochromatic light source is incident on a neutral density filter which decreases the light source's intensity by a factor of 3 but does not change its color. What happens to the energy of a photon in the less intense beam?

None of these

The photon energy decreases by a factor of 3.

The photon energy increases by a factor of 3.

The photon energy decreases by a factor of 9.

The photon energy decreases only slightly.

Explanation

The photon energy does not change since the color of light (and therefore the frequency of the light) does not change.

7

A car with mass is heading toward a wall at a speed $25\frac{m}{s}$ . What is the approximate de Broglie wavelength of the car?

$h = 6.626\times 10^{-34}~\text{J}\cdot\text{s}$

$1.33\times10^{-38}~\text{m}$

$2\times10^{-34}~\text{m}$

$80~\text{m}$

$0.0125~\text{m}$

$532\times10^{-9}~\text{m}$

Explanation

The de Broglie wavelength is given by $\lambda = \frac{h}{m v}$ .

8

Radiocarbon dating is a method that allows scientists to estimate the age of organisms after they have died. A fairly constant amount of radioactive $C^{14}$ remains in the animal while alive, but once it dies the $C^{14}$ degrades over time into $C^{12}$ . By measuring the relative amount of these two in a dead organism compared to a living one, along with knowing that the half-life of $C^{14}$ is years, it can be determined how long ago the organism died.

If a scientist finds that a fossil contains amount of the $C^{14}$ normally found in the atmosphere at that time, approximately how old is this fossil?

Explanation

In this question, we're given a brief description of radiocarbon dating. We're given the amount of $C^{14}$ that has been found in a fossil sample and we're asked to find the approximate age of the fossil.

First, let's briefly go over radiocarbon dating. This method essentially assumes that the amount of radioactive carbon within an organism remains fairly stable at any given time while the organism is alive. Moreover, this amount of radioactive carbon is related to the amount of radioactive carbon in the atmosphere. Once the organism dies, however, it ceases to gain any radioactive carbon; rather, the $C^{14}$ that was present now begins to decay into $C^{12}$ . Thus, by measuring the amount of $C^{14}$ in the organism and comparing it to the amount in the atmosphere, the age at which the organism died can be approximated.

Since we're dealing with the decay of $C^{14}$ , this is a radioactive decay problem. Recall that all radioactive decay reactions follow first-order rate kinetics. What this means is that the rate of decay is only dependent on the amount of radioactive material at any given instant. Hence, we can use the first-order rate equation.

$A_{t}=A_{0}e^{-kt}$

We can further rearrange this expression to isolate the variable for time.

$\frac{A_t}{A_0}=e^{-kt}$

$t=-\frac{ln(\frac{A_t}{A_0})}{k}$

In arriving at this expression, we see that we need to know the rate constant, , in order to solve for . To do this, we can use the equation that relates the half-life to the rate constant for a first-order process.

$t_\frac{1}{2}=\frac{ln(2)}{k}$

Rearranging, we can find the rate-constant.

$k=\frac{ln(2)}{t_\frac{1}{2}}$

$k=\frac{0.693}{5730: yr}=1.21\cdot 10^{-4}: yr^{-1}$

Now that we have the rate constant, we can plug this value into the previous expression to solve for .

$t=-\frac{ln(\frac{0.35}{1})}{1.21\cdot 10^{-4}: yr^{-1}}=8676:years$

9

A car with mass is heading toward a wall at a speed $25\frac{m}{s}$ . What is the approximate de Broglie wavelength of the car?

$h = 6.626\times 10^{-34}~\text{J}\cdot\text{s}$

$1.33\times10^{-38}~\text{m}$

$2\times10^{-34}~\text{m}$

$80~\text{m}$

$0.0125~\text{m}$

$532\times10^{-9}~\text{m}$

Explanation

The de Broglie wavelength is given by $\lambda = \frac{h}{m v}$ .

10

A laser outputs light of a single color with photon energy . A nonlinear crystal, which halves the wavelength of light, is placed in the path of the laser beam. What is the energy of the photons exiting the crystal?

Explanation

Halving the wavelength is equivalent to doubling the frequency since for a light wave the wavelength and frequency are related by $\lambda\nu = c$ , where is the speed of light. The energy of a photon is $E_\text{ph} = h\nu$ , where is Planck's constant, so the photon energy is doubled.

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