Question 1

What value of Z (atomic number) and A (mass number) result in the following β-decay? 
A) Z = 6; A = 12
B) Z = 5; A = 8
C) Z = 6; A = 11
D) Z = 8; A = 12
E) Z = 8; A = 11

Accepted Answer

A

Question 2

The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance. How much tritium will be left after 30 years?
A) 0.20 kg
B) 1.8 kg
C) 0.18 kg
D) 1.7 kg
E) 4.1 kg

Accepted Answer

After 30 years, which is approximately 2.44 half-lives (30 / 12.3), the amount of tritium left is calculated using the formula: initial amount * (1/2)^(number of half-lives). Thus, 10 * (1/2)^2.44 ≈ 1.8 kg.

Question 3

What value of Z (atomic number) and A (mass number) result in the following β-decay? 
A) Z = 5; A = 14
B) Z = 4; A = 10
C) Z = 6; A = 14
D) Z = 7; A = 14
E) Z = 7; A = 13

Accepted Answer

D

Question 4

How many radioactive atoms are present in a sample that has an activity of 0.5 μCi and a half-life of 10 years? (1 curie = 3.7 × 10¹⁰ decays/s) A) 9.5 × 10¹² atoms B) 8.4 × 10¹² atoms C) 7.3 × 10¹² atoms D) 6.5 × 10¹² atoms E) 2.7 × 10⁵ atoms

Accepted Answer

First, we need to convert the activity from microcuries to Curies:

0.5 μCi = 0.5 × 10^-6 Ci

Next, we can use the following equation to calculate the number of radioactive atoms present:

N = (A/λ) × (1 - e^(-λt))

where:
N = number of radioactive atoms
A = activity in decays/s (converted from Ci)
λ = decay constant ln2/t1/2 (where t1/2 is the half-life)
t = time elapsed

Let's plug in the values we know:

N = [(0.5 × 10^-6 Ci) / (ln2/10 yr)] × (1 - e^(-(ln2/10 yr) × 0 yr))

Simplifying this equation gives:

N = (0.5 × 10^-6 Ci) × (10 yr/ln2) = 8.38 × 10^12 radioactive atoms

Therefore, the answer is B: 8.4 × 10^12 atoms.

Question 5

The half-life of ¹³¹I is 8.04 days. Three days after it was prepared, its activity was 0.50 μCi. How many curies (in μCi) were initially prepared? A) .60 B) .70 C) .65 D) .55 E) .39

Accepted Answer

The answer of The half-life of ¹³¹I is 8.04 days....

Question 6

The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance. What will be the initial decay rate, at t = 0 (in decays/s)? A) 1.09 × 10¹⁴ B) 1.8 × 10−⁹ C) 5.6 × 10⁸ D) 3.6 × 10¹⁸ E) 3.6 × 10¹⁷

Accepted Answer

The answer of The isotope, tritium, has a half-life of...

Question 7

Find the binding energy per nucleon (in MeV/nucleon) of carbon-12. Assume: M_C = 12.000 000 u M _p = 1.007 825 u M _n = 1.008 665 u U = 1.66 × 10−²⁷ kg A) 1.2 B) 4.2 × 10−² C) 7.4 D) 7.7 E) 5.6

Accepted Answer

The binding energy of a nucleus is given by the formula

BE = [Z(Mp) + (A-Z)(Mn) - M]c^2

where BE is the binding energy, Z is the atomic number, A is the mass number, Mp is the mass of a proton, Mn is the mass of a neutron, M is the mass of the nucleus, and c is the speed of light.

Using the given data, we can calculate the mass of carbon-12:

M = 12u x (1.66 x 10^-27 kg/u) = 1.99 x 10^-25 kg

We can also calculate the number of protons and neutrons in carbon-12:

Z = 6
A-Z = 6

Next, we can calculate the total mass of the protons and neutrons in carbon-12:

Z(Mp) + (A-Z)(Mn) = 6(1.007825u) + 6(1.008665u) = 12.09888u

Substituting these values into the binding energy formula, we get:

BE = [6(1.007825u) + 6(1.008665u) - 12.000000u]c^2 = 92.162 MeV

The binding energy per nucleon is then:

BE/A = 92.162 MeV/12 = 7.68 MeV/nucleon

Rounding to the nearest tenth gives a final answer of 7.7 MeV/nucleon, which corresponds to choice D.

Question 8

Naturally radioactive nuclei can decay spontaneously by emitting the following particles:
A) helium nuclei, electrons, photons
B) electrons, neutrons, protons
C) helium nuclei, electrons, protons
D) electrons, neutrons, photons
E) quarks and leptons

Accepted Answer

Naturally radioactive nuclei can decay by emitting helium nuclei (alpha decay), electrons (beta decay), and photons (gamma decay).

Question 9

Find the ratio of the binding energy per nucleon for helium ( ) to uranium-238 ( ). Assume: M _p = 1.007 825 u M _n = 1.008 665 u M_He = 4.002 603 u M_U = 238.050 786 u U = 1.66 × 10−²⁷ kg A) 1.07 B) 0.934 C) 63.7 D) 1.6 × 10−² E) 3.24

Accepted Answer

The answer of Find the ratio of the binding energy...

Question 10

The radius of a nucleus of  (in fm) is
A) 15.4
B) 5.5
C) 12.8
D) 6.6
E) 4.9

Accepted Answer

The answer of The radius of a nucleus of ...

Question 11

44 g of petrified wood was found in a petrified forest. A sample showed a ¹⁴C activity of 100 decays/minute. How long has the tree been dead (in years)? (The half-life of carbon-14 is 5730 years and freshly cut wood contains 6.5 × 10¹⁰ atoms of ¹⁴C per gram.) A) 12300 B) 15600 C) 8500 D) 4700 E) 2400

Accepted Answer

The answer of 44 g of petrified wood was found...

Question 12

The ratio of the radius of a classical electron (r_e = k_ee²/m_ec² = 2.8 × 10−¹⁵ m) to the radius of a ⁴He nucleus (r = r₀A^1/3) is A) 2.0 B) 0.68 C) 1.5 D) 0.92 E) 2.4

Accepted Answer

The answer of The ratio of the radius of a...

Question 13

The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance. What will be the disintegration constant (in s−¹)? A) 5.6 × 10−² B) 5.6 × 10⁸ C) 3.2 × 10⁷ D) 1.8 × 10−⁹ E) 1.6 × 10⁶

Accepted Answer

The disintegration constant can be found using the formula: 
$\lambda = \frac{0.693}{t_{1/2}}$
Substituting the given values, we get:
$\lambda = \frac{0.693}{12.3	ext{ years}} \cdot \frac{1 	ext{ year}}{3.1536 \cdot 10^7 	ext{ s}} = 1.843 \cdot 10^{-19} 	ext{ s}^{-1}$
Therefore, the answer is D: 1.8 × 10−19 s−1.

Question 14

An alpha particle is emitted from a radioactive source with an energy of 5 MeV. How fast is it moving (in m/s)? (m = 4.002 603 u, 1 u = 1.66 × 10−²⁷ kg.) A) 2.4 × 10⁷ B) 1.6 × 10⁷ C) 3.7 × 10⁷ D) 4.6 × 10⁷ E) 2.1 × 10⁷

Accepted Answer

To find the velocity of the alpha particle, we can use the kinetic energy formula: $KE = \frac{1}{2}mv^2$, where $m$ is the mass of the alpha particle and $v$ is its velocity. First, convert the mass of the alpha particle from atomic mass units (u) to kilograms (kg) using the given conversion factor $1 u = 1.66 \times 10^{-27} kg$. The mass of the alpha particle is $4.002603 u \times 1.66 \times 10^{-27} kg/u = 6.6442 \times 10^{-27} kg$. The energy given is 5 MeV, which needs to be converted to joules (J). $1 eV = 1.602 \times 10^{-19} J$, so $5 MeV = 5 \times 10^6 eV = 5 \times 10^6 \times 1.602 \times 10^{-19} J = 8.01 \times 10^{-13} J$. Solving the kinetic energy formula for $v$, we get $v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2 \times 8.01 \times 10^{-13} J}{6.6442 \times 10^{-27} kg}} \approx 1.6 \times 10^{7} m/s$.

Question 15

The ratio of the density of a neutron (r = r₀A^1/3) to the density of a classical electron (r_e = ke²/m_ec² = 2.8 × 10−¹⁵ m) is A) 4.3 × 10² B) 2.3 × 10⁴ C) 1.4 × 10² D) 6.9 × 10−³ E) 4.3 × 10³

Accepted Answer

The density of a neutron is significantly higher than the density of a classical electron due to the compact mass of a neutron compared to the relatively small mass and larger volume associated with the classical electron's theoretical space. The calculation involves comparing the mass and volume (or space) occupied by both particles, leading to a substantial difference in density.

Question 16

What value of Z (atomic number) and A (mass number) result in the following alpha decay? 
A) Z = 92; A = 238
B) Z = 91; A = 238
C) Z = 90; A = 234
D) Z = 93; A = 238
E) Z = 88; A = 236

Accepted Answer

In alpha decay, a nucleus emits an alpha particle, which is composed of 2 protons and 2 neutrons (i.e. a helium nucleus). 
The given alpha decay can be written as: 
             11eb6adc_ed7f_b1c5_b0cf_1d84624c23fc_TB8245_11 --> Z-2, A-4 + 2He4
Here, Z and A are the atomic number and mass number of the parent nucleus, respectively. 
Comparing the given alpha decay with the general form, we can determine the values of Z and A for the parent nucleus: 
             (Z, A) --> (Z-2, A-4)
Therefore, the parent nucleus has Z = 90 (i.e. 92-2) and A = 234 (i.e. 238-4). 
Among the answer choices, only option C has Z = 90 and A = 234.

Question 17

For large mass number nuclei which are stable, the ratio of protons to neutrons is
A) equal to 1
B) greater than 1
C) less than 1
D) unrelated to the stability of nuclei
E) almost 2 to 1

Accepted Answer

For large mass number nuclei which are stable, the ratio of protons to neutrons is typically less than 1. This is because as the number of protons increases, the electrostatic repulsion between protons increases, making it more difficult to hold the nucleus together. Therefore, a larger number of neutrons is needed to provide the strong nuclear force necessary to overcome the electrostatic repulsion and hold the nucleus together. This is known as the neutron-to-proton ratio and is a key factor in determining the stability of nuclei.

Question 18

Two isotopes of uranium have the same
A) mass number
B) neutron number
C) atomic number
D) nucleon number
E) nucleon number and neutron number

Accepted Answer

Isotopes of an element, including uranium, have the same atomic number because they have the same number of protons in their nuclei. The atomic number defines the element.

Question 19

Calculate the binding energy per nucleon (MeV/nucleon) for tritium, ( ) a radioactive isotope of hydrogen. Assume: M _p = 1.007 825 u M _n = 1.008 665 u M _t = 3.016 05 u U = 1.66 × 10−²⁷ kg A) 2.8 B) 3.1 C) 1.0 D) 8.5 E) 2.1

Accepted Answer

The answer of Calculate the binding energy per nucleon (MeV/nucleon)...

Question 20

Find the binding energy (in MeV) of carbon-12. Assume: M_C = 12.000 000 u M _p = 1.007 825 u M _n = 1.008 665 u U = 1.66 × 10−²⁷ kg A) 14.8 B) 0.511 C) 9.11 D) 92.3 E) 46.2

Accepted Answer

The binding energy of a nucleus can be calculated using the mass defect, which is the difference between the mass of the nucleus and the sum of the masses of its protons and neutrons. For carbon-12 ($^{12}C$), which has 6 protons and 6 neutrons:Mass of protons = $6 \times 1.007825 \, u = 6.04695 \, u$Mass of neutrons = $6 \times 1.008665 \, u = 6.05199 \, u$Total mass of protons and neutrons = $6.04695 \, u + 6.05199 \, u = 12.09894 \, u$Mass defect = $12.09894 \, u - 12.000000 \, u = 0.09894 \, u$To convert the mass defect into energy (MeV), we use Einstein's equation $E = mc^2$, where $m$ is the mass in kilograms, $c$ is the speed of light ($3.00 \times 10^8 \, m/s$), and $E$ is the energy in joules. However, in nuclear physics, it's common to use atomic mass units (u) and directly convert to MeV using the conversion factor $1 \, u = 931.5 \, MeV/c^2$:Binding energy = $0.09894 \, u \times 931.5 \, MeV/u = 92.16 \, MeV$Therefore, the closest answer is 92.3 MeV.

Quiz 44: Nuclear Structure

What value of Z (atomic number) and A (mass number) result in the following β-decay?

The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance. How much tritium will be left after 30 years?

What value of Z (atomic number) and A (mass number) result in the following β-decay?

How many radioactive atoms are present in a sample that has an activity of 0.5 μCi and a half-life of 10 years? (1 curie = 3.7 × 10¹⁰ decays/s)

The half-life of ¹³¹I is 8.04 days. Three days after it was prepared, its activity was 0.50 μCi. How many curies (in μCi) were initially prepared?

The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance. What will be the initial decay rate, at t = 0 (in decays/s) ?

Find the binding energy per nucleon (in MeV/nucleon) of carbon-12. Assume: M_C = 12.000 000 u M _p = 1.007 825 u M _n = 1.008 665 u U = 1.66 × 10−²⁷ kg

Naturally radioactive nuclei can decay spontaneously by emitting the following particles:

Find the ratio of the binding energy per nucleon for helium ( ) to uranium-238 ( ) . Assume: M _p = 1.007 825 u M _n = 1.008 665 u M_He = 4.002 603 u M_U = 238.050 786 u U = 1.66 × 10−²⁷ kg

The radius of a nucleus of (in fm) is

44 g of petrified wood was found in a petrified forest. A sample showed a ¹⁴C activity of 100 decays/minute. How long has the tree been dead (in years) ? (The half-life of carbon-14 is 5730 years and freshly cut wood contains 6.5 × 10¹⁰ atoms of ¹⁴C per gram.)

The ratio of the radius of a classical electron (r_e = k_ee²/m_ec² = 2.8 × 10−¹⁵ m) to the radius of a ⁴He nucleus (r = r₀A^1/3) is

The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance. What will be the disintegration constant (in s−¹) ?

An alpha particle is emitted from a radioactive source with an energy of 5 MeV. How fast is it moving (in m/s) ? (m = 4.002 603 u, 1 u = 1.66 × 10−²⁷ kg.)

The ratio of the density of a neutron (r = r₀A^1/3) to the density of a classical electron (r_e = ke²/m_ec² = 2.8 × 10−¹⁵ m) is

What value of Z (atomic number) and A (mass number) result in the following alpha decay?

For large mass number nuclei which are stable, the ratio of protons to neutrons is

Two isotopes of uranium have the same

Calculate the binding energy per nucleon (MeV/nucleon) for tritium, ( ) a radioactive isotope of hydrogen. Assume: M _p = 1.007 825 u M _n = 1.008 665 u M _t = 3.016 05 u U = 1.66 × 10−²⁷ kg

Find the binding energy (in MeV) of carbon-12. Assume: M_C = 12.000 000 u M _p = 1.007 825 u M _n = 1.008 665 u U = 1.66 × 10−²⁷ kg

Quiz 44: Nuclear Structure

What value of Z (atomic number) and A (mass number) result in the following β-decay?

The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance. How much tritium will be left after 30 years?

What value of Z (atomic number) and A (mass number) result in the following β-decay?

How many radioactive atoms are present in a sample that has an activity of 0.5 μCi and a half-life of 10 years? (1 curie = 3.7 × 1010 decays/s)

The half-life of 131I is 8.04 days. Three days after it was prepared, its activity was 0.50 μCi. How many curies (in μCi) were initially prepared?

The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance. What will be the initial decay rate, at t = 0 (in decays/s) ?

Find the binding energy per nucleon (in MeV/nucleon) of carbon-12. Assume: MC = 12.000 000 u M p = 1.007 825 u M n = 1.008 665 u U = 1.66 × 10−27 kg

Naturally radioactive nuclei can decay spontaneously by emitting the following particles:

Find the ratio of the binding energy per nucleon for helium ( ) to uranium-238 ( ) . Assume: M p = 1.007 825 u M n = 1.008 665 u MHe = 4.002 603 u MU = 238.050 786 u U = 1.66 × 10−27 kg

The radius of a nucleus of (in fm) is

44 g of petrified wood was found in a petrified forest. A sample showed a 14C activity of 100 decays/minute. How long has the tree been dead (in years) ? (The half-life of carbon-14 is 5730 years and freshly cut wood contains 6.5 × 1010 atoms of 14C per gram.)

The ratio of the radius of a classical electron (re = kee2/mec2 = 2.8 × 10−15 m) to the radius of a 4He nucleus (r = r0A1/3) is

The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance. What will be the disintegration constant (in s−1) ?

An alpha particle is emitted from a radioactive source with an energy of 5 MeV. How fast is it moving (in m/s) ? (m = 4.002 603 u, 1 u = 1.66 × 10−27 kg.)

The ratio of the density of a neutron (r = r0A1/3) to the density of a classical electron (re = ke2/mec2 = 2.8 × 10−15 m) is

What value of Z (atomic number) and A (mass number) result in the following alpha decay?

For large mass number nuclei which are stable, the ratio of protons to neutrons is

Two isotopes of uranium have the same

Calculate the binding energy per nucleon (MeV/nucleon) for tritium, ( ) a radioactive isotope of hydrogen. Assume: M p = 1.007 825 u M n = 1.008 665 u M t = 3.016 05 u U = 1.66 × 10−27 kg

Find the binding energy (in MeV) of carbon-12. Assume: MC = 12.000 000 u M p = 1.007 825 u M n = 1.008 665 u U = 1.66 × 10−27 kg

How many radioactive atoms are present in a sample that has an activity of 0.5 μCi and a half-life of 10 years? (1 curie = 3.7 × 10¹⁰ decays/s)

The half-life of ¹³¹I is 8.04 days. Three days after it was prepared, its activity was 0.50 μCi. How many curies (in μCi) were initially prepared?

Find the binding energy per nucleon (in MeV/nucleon) of carbon-12. Assume: M_C = 12.000 000 u M _p = 1.007 825 u M _n = 1.008 665 u U = 1.66 × 10−²⁷ kg

Find the ratio of the binding energy per nucleon for helium ( ) to uranium-238 ( ) . Assume: M _p = 1.007 825 u M _n = 1.008 665 u M_He = 4.002 603 u M_U = 238.050 786 u U = 1.66 × 10−²⁷ kg

44 g of petrified wood was found in a petrified forest. A sample showed a ¹⁴C activity of 100 decays/minute. How long has the tree been dead (in years) ? (The half-life of carbon-14 is 5730 years and freshly cut wood contains 6.5 × 10¹⁰ atoms of ¹⁴C per gram.)

The ratio of the radius of a classical electron (r_e = k_ee²/m_ec² = 2.8 × 10−¹⁵ m) to the radius of a ⁴He nucleus (r = r₀A^1/3) is

The isotope, tritium, has a half-life of 12.3 years. Assume we have 10 kg of the substance. What will be the disintegration constant (in s−¹) ?

An alpha particle is emitted from a radioactive source with an energy of 5 MeV. How fast is it moving (in m/s) ? (m = 4.002 603 u, 1 u = 1.66 × 10−²⁷ kg.)

The ratio of the density of a neutron (r = r₀A^1/3) to the density of a classical electron (r_e = ke²/m_ec² = 2.8 × 10−¹⁵ m) is

Calculate the binding energy per nucleon (MeV/nucleon) for tritium, ( ) a radioactive isotope of hydrogen. Assume: M _p = 1.007 825 u M _n = 1.008 665 u M _t = 3.016 05 u U = 1.66 × 10−²⁷ kg

Find the binding energy (in MeV) of carbon-12. Assume: M_C = 12.000 000 u M _p = 1.007 825 u M _n = 1.008 665 u U = 1.66 × 10−²⁷ kg