Question 1

It is often possible to use atomic masses when calculating the binding energy of a nucleus.This is not true for calculating the Q value for the e⁺ decay process since

Accepted Answer

The Q value for a decay process involves the difference in mass-energy between the initial and final states. In the case of electron capture (e⁺ decay process), the mass of the electron is crucial because it is directly involved in the reaction, unlike in many nuclear reactions where electron masses can be neglected due to their relatively small mass compared to nucleons. The electron is not merely a spectator but participates in the reaction, thus its mass does not cancel out and must be considered in the calculation.

Question 2

How fast must two deuterium atoms be moving so they can overcome the Coulomb force of repulsion,and attain the necessary 10 ^- ¹⁴ m for fusion? (m( ${ } _ { 1 } ^ { 2 } \mathrm { H }$ )= 2.014 1 u)

Accepted Answer

C

Question 3

What energy is needed (in MeV)so two deuterium atoms moving together will reach the necessary 10 ^- ¹⁴ m for fusion?

Accepted Answer

B

Question 4

Two nuclei may have equal Z,but different A,because they contain

Accepted Answer

The letter "Z" refers to the atomic number, which is equivalent to the number of protons in the nucleus. Therefore, if two nuclei have equal Z, they have an equal number of protons. However, the letter "A" refers to the atomic mass, which includes the number of protons and neutrons. Therefore, if two nuclei have different A, it means they have different numbers of neutrons. Choice A is incorrect because if two nuclei have equal numbers of protons and neutrons, they would have equal A. Choice C is incorrect because if two nuclei have equal numbers of neutrons but different numbers of protons, they would have different Z. Choice D is incorrect because it is too broad and could encompass all possible scenarios. Choice E is incorrect because electrons are not part of the nucleus.

Question 5

Two nuclei which share the same mass number A always are (Hint: Eliminate any wrong answers. )

Accepted Answer

The answer of Two nuclei which share the same mass...

Question 6

Find the number of nuclei per unit volume (n = nuclei/cm³)for lead. atomic weight = 202.7 Density = 11.5 g/cm³ Avogadro's number = 6.02 *10²³

Accepted Answer

The answer of Find the number of nuclei per unit...

Question 7

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

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 8

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

Accepted Answer

The binding energy of a nucleus is given by
E = [Zmp + (A − Z)mn − M]c^2
where Z is the atomic number, A is the mass number, mp and mn are the masses of the proton and neutron respectively and M is the mass of the nucleus.
For carbon-12:
Z = 6
A = 12
mp = 1.007825 u
mn = 1.008665 u
M = 12.000000 u
Substituting the values, we get
E = [6(1.007825) + 6(1.008665) - 12.000000]c^2
= -92.16 MeV
The binding energy per nucleon is given by
BE/A = E/A
Substituting the values, we get
BE/A = -92.16/12
= -7.68 MeV/nucleon
Taking the absolute value, we get
BE/A = 7.68 MeV/nucleon
Since this is closest to choice D, the answer is D.

Question 9

Two nuclei which share the same atomic number Z always are

Accepted Answer

Nuclei with the same atomic number Z are isotopes of each other, meaning they have the same number of protons but can have a different number of neutrons. Isotopes can be stable or unstable, depending on their neutron-to-proton ratio. The term "isobar" refers to nuclei with the same mass number (total number of protons and neutrons), while the term "radioactive" refers to nuclei that undergo spontaneous decay. So, the best choice is C, as it correctly identifies these nuclei as isotopes of each other.

Question 10

Calculate the binding energy per nucleon (MeV/nucleon)for tritium,( ${ } _ { 1 } ^ { 3 } \mathrm { H }$ )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

Accepted Answer

The binding energy per nucleon is calculated using the formula: $ BE/A = \frac{(Z \cdot M_p + N \cdot M_n - M_t) \cdot 931.5}{A} $. For tritium, Z=1, N=2, A=3. Substituting the given masses, the binding energy per nucleon is approximately 2.8 MeV/nucleon.

Question 11

It is often possible to use the atomic masses when calculating the binding energy of a nucleus.The reason for this is

Accepted Answer

The binding energy of a nucleus is primarily due to the strong nuclear force, which acts only between nucleons (protons and neutrons) in the nucleus. Electrons do not participate in this force and their masses cancel out when calculating the binding energy. Therefore, it is often possible to use the atomic masses (which include the masses of both protons and neutrons) when calculating the binding energy of a nucleus.

Question 12

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

Accepted Answer

The ratio of density of a neutron to a classical electron is given by r/r_e. Substituting the values of the constants, we get: r/r_e = k*(m_e/m_n)*(c^2)/(h^2) r/r_e = (2.8*10^-10)*(9.11*10^-31/1.67*10^-27)*((3*10^8)^2)/((6.626*10^-34)^2) r/r_e = 2.3*10^12 Therefore, the answer is B, 2.3*10^12 or 2.3 F1F1F1S1F1F1F10 10⁴.

Question 13

Heavy nuclei are unstable because

Accepted Answer

Heavy nuclei have a large number of protons and neutrons, which leads to a strong Coulomb repulsive force between the protons. The nuclear force, which is responsible for holding the nucleus together, is strong enough to overcome this repulsion, but only at distances closer than about 2 femtometers (fm). At greater distances, the force falls off rapidly, leading to instability. This is why many heavy nuclei undergo radioactive decay, in which they emit particles to become more stable.

Question 14

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

Accepted Answer

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

Question 15

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

Accepted Answer

The answer of Find the binding energy (in MeV)of carbon-12....

Question 16

The radius of a nucleus of ${ } _ { 67 } ^ { 165 } \mathrm { H o}$ (in fm)is

Accepted Answer

The radius of a nucleus can be estimated using the formula $ R = R_0 A^{1/3} $, where $ R_0 \approx 1.2 $ fm and $ A $ is the mass number. For ${ } _ { 67 } ^ { 165 } \mathrm { H o}$, $ A = 165 $. Calculating gives $ R \approx 1.2 \times 165^{1/3} \approx 6.6 $ fm.

Question 17

Rutherford's experiment,in which he fired alpha particles of 7.7 MeV kinetic energy at a thin gold foil,showed that nuclei were very much smaller than the size of an atom because

Accepted Answer

Rutherford's experiment showed that some alpha particles were deflected backwards, indicating that they had encountered a densely packed, positively charged region within the atom - which we now know to be the nucleus. The fact that only a tiny fraction of alpha particles were deflected in this way suggested that the nucleus must be very small compared to the size of an atom. Choice A is partially correct, as some alpha particles did pass through the foil undeflected, but this does not provide evidence for the small size of the nucleus. Choices C, D, and E are all incorrect, as there is no evidence from Rutherford's experiment that alpha particles were captured by gold nuclei, could not get closer than a certain distance to the nuclei, or split into deuterium nuclei.

Question 18

The radius of an approximately spherical nucleus is given by r =

Accepted Answer

The radius of a nucleus is given by the formula r = r0A^(1/3), where r0 is a constant and A is the mass number of the nucleus. We can rewrite this formula in terms of the number of protons Z as r = r0(ZA)^(1/3). Comparing this formula with the options given, we see that option D matches this formula exactly, with r0 = r₀A and Z = 2. This suggests that option D is the correct choice for the radius of a helium nucleus. Option A contains an extra factor of Z, which would make the radius too large. Option B contains an incorrect power of Z (11/3 instead of 1/3) and also an extra factor of 3 in the denominator, which would make the radius too small. Option C is the same as option A but with A instead of Z, which is also incorrect. Option E has an extra factor of (AF1F1F1S1F1F1F10 Z) in the middle, which is not part of the formula for r.

Question 19

Two isotopes of uranium have the same

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 20

Find the ratio of the binding energy per nucleon for helium ( ${ } _ { 2 } ^ { 4 } \mathrm { He }$ )to uranium-238 ( ${ } _ { 92 } ^ { 238 } \mathrm { U }$ ). 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

Accepted Answer

The binding energy per nucleon can be calculated using the formula E/A, where E is the binding energy and A is the number of nucleons. Using the given data, the binding energy per nucleon for helium is:

E He = 2 × 4.002603 u - 4.030387 u = 28.30 MeV
E/A He = 28.30 MeV / 2 = 14.15 MeV/nucleon

Similarly, the binding energy per nucleon for uranium-238 is:

E U = 238 × 238.050786 u - 238.908931 u = 1785.93 MeV
E/A U = 1785.93 MeV / 238 = 7.49 MeV/nucleon

The ratio of E/A for helium to uranium-238 is:

(E/A He) / (E/A U) = (14.15 MeV/nucleon) / (7.49 MeV/nucleon) = 1.888

Therefore, the closest answer choice is B) 0.934, which is the inverse of the calculated ratio.

Quiz 30: Nuclear Physics

It is often possible to use atomic masses when calculating the binding energy of a nucleus.This is not true for calculating the Q value for the e⁺ decay process since

How fast must two deuterium atoms be moving so they can overcome the Coulomb force of repulsion,and attain the necessary 10 ^- ¹⁴ m for fusion? (m( ${ } _ { 1 } ^ { 2 } \mathrm { H }$ ) = 2.014 1 u)

What energy is needed (in MeV) so two deuterium atoms moving together will reach the necessary 10 ^- ¹⁴ m for fusion?

Two nuclei may have equal Z,but different A,because they contain

Two nuclei which share the same mass number A always are (Hint: Eliminate any wrong answers. )

Find the number of nuclei per unit volume (n = nuclei/cm³) for lead. atomic weight = 202.7 Density = 11.5 g/cm³ Avogadro's number = 6.02 *10²³

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

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

Two nuclei which share the same atomic number Z always are

Calculate the binding energy per nucleon (MeV/nucleon) for tritium,( ${ } _ { 1 } ^ { 3 } \mathrm { H }$ ) 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

It is often possible to use the atomic masses when calculating the binding energy of a nucleus.The reason for this is

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

Heavy nuclei are unstable because

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

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

The radius of a nucleus of ${ } _ { 67 } ^ { 165 } \mathrm { H o}$ (in fm) is

Rutherford's experiment,in which he fired alpha particles of 7.7 MeV kinetic energy at a thin gold foil,showed that nuclei were very much smaller than the size of an atom because

The radius of an approximately spherical nucleus is given by r =

Two isotopes of uranium have the same

Find the ratio of the binding energy per nucleon for helium ( ${ } _ { 2 } ^ { 4 } \mathrm { He }$ ) to uranium-238 ( ${ } _ { 92 } ^ { 238 } \mathrm { U }$ ) . 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

Quiz 30: Nuclear Physics

It is often possible to use atomic masses when calculating the binding energy of a nucleus.This is not true for calculating the Q value for the e+ decay process since

How fast must two deuterium atoms be moving so they can overcome the Coulomb force of repulsion,and attain the necessary 10 - 14 m for fusion? (m( 12H{ } _ { 1 } ^ { 2 } \mathrm { H }12​H ) = 2.014 1 u)

What energy is needed (in MeV) so two deuterium atoms moving together will reach the necessary 10 - 14 m for fusion?

Two nuclei may have equal Z,but different A,because they contain

Two nuclei which share the same mass number A always are (Hint: Eliminate any wrong answers. )

Find the number of nuclei per unit volume (n = nuclei/cm3) for lead. atomic weight = 202.7 Density = 11.5 g/cm3 Avogadro's number = 6.02 *1023

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

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

Two nuclei which share the same atomic number Z always are

Calculate the binding energy per nucleon (MeV/nucleon) for tritium,( 13H{ } _ { 1 } ^ { 3 } \mathrm { H }13​H ) 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

It is often possible to use the atomic masses when calculating the binding energy of a nucleus.The reason for this is

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

Heavy nuclei are unstable because

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

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

The radius of a nucleus of 67165Ho{ } _ { 67 } ^ { 165 } \mathrm { H o}67165​Ho (in fm) is

Rutherford's experiment,in which he fired alpha particles of 7.7 MeV kinetic energy at a thin gold foil,showed that nuclei were very much smaller than the size of an atom because

The radius of an approximately spherical nucleus is given by r =

Two isotopes of uranium have the same

Find the ratio of the binding energy per nucleon for helium ( 24He{ } _ { 2 } ^ { 4 } \mathrm { He }24​He ) to uranium-238 ( 92238U{ } _ { 92 } ^ { 238 } \mathrm { U }92238​U ) . 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

It is often possible to use atomic masses when calculating the binding energy of a nucleus.This is not true for calculating the Q value for the e⁺ decay process since

How fast must two deuterium atoms be moving so they can overcome the Coulomb force of repulsion,and attain the necessary 10 ^- ¹⁴ m for fusion? (m( ${ } _ { 1 } ^ { 2 } \mathrm { H }$ ) = 2.014 1 u)

What energy is needed (in MeV) so two deuterium atoms moving together will reach the necessary 10 ^- ¹⁴ m for fusion?

Find the number of nuclei per unit volume (n = nuclei/cm³) for lead. atomic weight = 202.7 Density = 11.5 g/cm³ Avogadro's number = 6.02 *10²³

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

Calculate the binding energy per nucleon (MeV/nucleon) for tritium,( ${ } _ { 1 } ^ { 3 } \mathrm { H }$ ) 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

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

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

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

The radius of a nucleus of ${ } _ { 67 } ^ { 165 } \mathrm { H o}$ (in fm) is

Find the ratio of the binding energy per nucleon for helium ( ${ } _ { 2 } ^ { 4 } \mathrm { He }$ ) to uranium-238 ( ${ } _ { 92 } ^ { 238 } \mathrm { U }$ ) . 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