A. Incorrect! Do not confuse Nucleus, Neutron and Nucleon. B. Incorrect! Nucleon is the name given to the two particles that make up the nucleus.

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1 AP Physics - Problem Drill 24: Nuclear Physics 1. Identify what is being described in each of these statements. Question 01 (1) It is held together by the extremely short range Strong force. (2) The magnitude of its charge is the same as the electron. (3) Each isotope of a particular element has a different number of these. (A) Neutron (2) Proton (3) Nucleus (B) Proton (2) Neutron (3) Nucleon (C) Nucleus (2) Proton (3) Neutron (D) Proton (2) Nucleus (3) Neutron (E) Nucleus (2) Proton (3) Nucleon Do not confuse Nucleus, Neutron and Nucleon. Nucleon is the name given to the two particles that make up the nucleus. C. Correct! The nucleus contains protons and neutrons, the force that hold these nucleons together in the nucleus is the Strong Nuclear Force this force only acts over short distances (<10-15 m). The magnitude of the charge on the proton is the same as the electron but has the opposite sign. Isotopes of the same element contain the same number of protons but different number of neutrons. The Nucleon is the name given to the two particles that make up the nucleus. Nucleon is the name given to the two particles that make up the nucleus. The nucleus contains protons and neutrons, the force that hold these nucleons together in the nucleus is the Strong Nuclear Force this force only acts over short distances (<10-15 m). The magnitude of the charge on the proton is the same as the electron but has the opposite sign. Isotopes of the same element contain the same number of protons but different number of neutrons. The term nucleon is given the two particles that make up the nucleus namely the proton and neutron. The correct answer is (C).

2 Question No. 2 of How many protons and neutrons would be found in an atom of Iron with an atomic mass of 5 and an atomic number of 2? Question 02 (A) 5 protons and 2 neutrons (B) 2 protons and 5 neutrons (C) 2 protons and 30 neutrons (D) 30 protons and 2 neutrons (E) 5 protons and 30 neutrons 5 protons would require an atomic number of 5 and 2 neutrons means that the atomic mass would be or 82. This does NOT meet the situation described in the question, where the atomic number was 2 and the atomic mass is 5. 2 protons would require an atomic number of 2 and 5 neutrons means that the atomic mass would still be or 82. This does NOT meet the situation described in the question, where the atomic number was 2 and the atomic mass is 5. C. Correct! 2 protons would require an atomic number of 2 and 30 neutrons means that the atomic mass would be or 5. This DOES meet the situation described in the question, where the atomic number was 2 and the atomic mass is protons would require an atomic number of 30 and 2 neutrons means that the atomic mass would be or 5. This does NOT meet the situation described in the question, where the atomic number was 2 and the atomic mass is 5. 5 protons would require an atomic number of 5 and 30 neutrons requires an atomic mass of 5+30 = 8. This does NOT meet the situation described in the question, where the atomic number was 2 and the atomic mass is 5. Atomic number is equal to the number of protons in the nucleus and has the symbol Z Atomic mass number (atomic mass or mass number) is the total number of protons and neutrons in the nucleus. To find the number of neutrons subtract the atomic number from the atomic mass number. For an element with 2 protons the atomic number will be 2. If the atomic mass of this element is 5 then the number of neutrons is 5-2 = 30 The correct answer is (C).

3 Question No. 3 of A friend has asked you what binding energy is and how to calculate it. Which of these would not be part of your explanation? Question 03 (A) The total mass of a stable nucleus is always less than the sum of the masses of the individual protons and neutrons. (B) The missing mass has gone into energy that holds the nucleus together; this is what we call the total binding energy of the nucleus. (C) Binding energy is the energy that is released when the nucleus splits apart. (D) To find the value for binding energy we find the total mass of the protons and neutrons in an atom of the element, and then we subtract the actual measured mass of an atom. Remembering to take into account that this includes the electrons. (E) We must convert the atomic mass units into kilograms and then use the equation E = mc 2 to convert to energy in Joules. It you sum the mass of all the protons and neutrons in the nucleus, and compare this to actual mass value of nucleus you will find there is a discrepancy. This missing mass is transformed into energy that holds the nucleus together. C. Correct! Binding energy is the energy that is needed to break the nucleus apart. It is missing mass (energy) that has to be put back. This is the method to use. The units of mass are typically given in unified atomic mass units. Since the actual mass of the atom includes the electrons we have to subtract the electrons mass from the actual mass to find the mass of the nucleus. Remember the number of electrons is the same as the number of protons. We have to use this equation to find how much energy the missing mass is equivalent to. It you sum the mass of all the protons and neutrons in the nucleus, and compare this to actual mass value of nucleus you will find there is a discrepancy. This missing mass is transformed into energy that holds the nucleus together. Binding energy is the energy that is needed to break the nucleus apart. It is missing mass (energy) that has to be put back. To find the value for binding energy we find the total mass of the protons and neutrons in an atom of the element, and then we subtract the actual measured mass of an atom. Since the actual mass of the atom includes the electrons we have to subtract the electrons mass from the actual mass to find the mass of the nucleus. Remember the number of electrons is the same as the number of protons. The final steps in the calculation are, convert the atomic mass units into kilograms and then use the equation E = mc 2 to convert to energy in Joules. The correct answer is (C).

4 Question No. 4 of We have found the mass defect for an atom to be u. What is the binding energy for that atom in Joules? Question 04 (A) 5.98 x J (B) 4.53 x J (C) 1.99 x J (D).4 x J (E) 0.04 J A. Correct! This answer is a result of properly converting unified atomic mass units (u) to kilograms (kg) and then using E=mc 2 where we know m and c is the speed of light and calculate E. Do not assume the binding energy is always the same as the one calculated in the tutorial. C. Incorrect! Check that you squared the speed of light in the calculation. Do not forget to convert unified atomic mass units (u) to kilograms (kg). You need to convert from convert unified atomic mass units (u) to kilograms (kg) and then from kilograms to Joules using E=mc 2 where we know m and c is the speed of light and calculate E. Known: Mass in atomic mass units, 0.04 u Unknown: Equivalent energy, E =? Define: Output: First convert from u to kg, 1 u = 1.05 x kg Then use E=mc 2 c the speed of light 3.0 x 10 8 m/s kg u =.42 x 10 kg 1 u E=mc E = kg ( m / s) = J Substantiate: Units are correct, sig figs are correct, Magnitude is correct, The correct answer is (A).

5 Question No. 5 of 10 as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed. 5. What is the binding energy of an atom of 12 C where the measured mass of an atom is ? Hint the mass of 1 1H is u, the mass of neutron is u and 1 u =931 MeV. Question 05 (A) 0.10 MeV (B) 5.72 GeV (C) MeV (D) 0 (E) 92.2 MeV Don t forget to convert from mass to energy. Remember that the atomic mass number is the number of protons and neutrons. C. Incorrect! You do not need to subtract the mass of the electrons, since the mass of the Hydrogen takes this into account. There must be some energy holding the nucleus together, otherwise it would just fall apart. E. Correct! The number of protons and neutrons is ; there must be an equal number of electrons as protons. Calculate the expected mass using x the mass of the hydrogen (which has 1 e and 1 p) and the mass of neutrons. Subtract the actual mass which includes the electrons. Convert the mass to MeV. Known: Actual mass of 12 C M a= Mass of 1 1 H = u Mass of 1 neutron = u 1 u = 931 MeV Unknown: Binding energy of 12 C =? MeV Define: There are a number of methods that you could use but given the hint the following method is the straightest forward. Step 1 Use the nuclear symbol to find the number of protons and number of neutrons in 12 C. The top number is the atomic mass, A= protons + neutrons =12 Bottom number is the atomic number, Z = protons = Number of neutrons = Z-A = 12 = Step 2 Calculate the expected mass of the carbon atom. We are given the mass of hydrogen 1 1H, which has 1 proton, and therefore 1 electron and 0 neutrons. We can calculate the expected value of the mass of the protons and electrons in the carbon atom. Expected mass of protons and electrons, M p+e = x 1 1 H Expected mass of neutron, M n = x u Step 3 Subtract the actual mass, which includes electrons, from the expected mass Missing mass, M m = (M p+e + M n ) - M a Step 4 Next convert missing mass to binding energy 931 MeV Binding Energy = Mm 1 u Output: M p+e = x u =.048 u M n = x u =.0522 u Expected mass = u Actual mass = u Missing mass = u Binding energy = u x 931 MeV/u = MeV = 92.2 MeV to 3 sig fig Substantiate: Units are correct, sig figs are correct, magnitude is reasonable The correct answer is (E).

6 Question No. of 10 as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.. Which of these is an incorrect statement? Question 0 (A) The isotopes of an element will all contain the same number of neutrons. (B) The isotopes of an element do not occur in equal amounts in nature. (C) Isotopes have the same atomic number but different atomic mass. (D) Isotope may be stable or radioactive. (E) The chemical properties for isotopes of the element are for the most part the same. A. Correct! An isotope of an element contains the same number of protons but a different number of neutrons. For example carbon has these isotopes C C C C C The atomic number, protons, is the same but the atomic mass is different, so the number of neutrons is different. Different isotopes of the same element do not occur at the same level of abundance, for example 98.9% of naturally occurring carbon is C the other 1.1% is C.The other isotopes of carbon are not stable, 14 C, occurs naturally at one part in a trillion. C. Incorrect! An isotope of an element contains the same number of protons but a different number of neutrons. For example carbon has these isotopes C C C C C The atomic number is the number of protons, and is the same, but the atomic mass is different, indicating the number of neutrons is different. Some isotopes are stable, meaning that they do not decay or disintegrate, other isotopes are unstable; in which case they will emit radiation in order to become stable. For example 12 C and C are both stable, while C, is unstable and undergoes beta decay, which releases an electron and neutrino to become stable 14 7 N The addition of neutrons to the nucleus has very little effect on the chemical properties of the element. Chemical properties are for the most part dependent on the number of electrons, and since the isotopes contain the same number of electrons they have the same chemical properties. An isotope of an element contains the same number of protons but a different number of neutrons. For example carbon has these isotopes C C C C C The atomic number, protons, is the same but the atomic mass is different, so the number of neutrons is different. Different isotopes of the same element do not occur at the same level of abundance, for example 98.9% of naturally occurring carbon is C the other 1.1% is C.The other isotopes of carbon are not stable, 14 C, occurs naturally at one part in a trillion. Some isotopes are stable, meaning that they do not decay or disintegrate, do not decay or disintegrate, other isotopes are unstable, in which case they will emit radiation in order to become stable. For example C and C are both stable, while C, is unstable and undergoes beta decay, which releases an electron and neutrino to become stable 14 7 N The addition of neutrons to the nucleus has very little effect on the chemical properties of the element. Chemical properties are for the most part dependent on the number of electrons, and since the isotopes contain the same number of electrons they have the same chemical properties. The correct answer is (A).

7 Question No. 7 of An atom undergoes radioactive decay, but both the atomic number and the atomic mass remain the same. Which of the following was probably the type of decay occurred? Question 07 (A) Alpha decay (B) Beta decay (C) Gamma decay (D) Radioactive decay (E) It is not possible to determine. Alpha decay results in the expulsion of a particle of 2 protons and 2 neutrons. This changes the nucleus of the original into that of another element with an atomic number that is two greater than the original. Beta decay on results in a neutron converting into a proton as it emits an electron (beta particle). This changes the nucleus of the original into that of another element with an atomic number one greater than the original. C. Correct! Gamma decay is essentially a packet of energy (photon) emitted by the nucleus. When this happens, neither the number of protons or neutrons changes. In turn, the atomic number and atomic mass remains unaffected. Radioactive decay is how we refer to the different types of processes that can occur when an unstable nucleus emits energy or particles. Each type of decay has a particular characteristic; when different particles are emitted the effect on the element will be different. Gamma decay occurs when a high energy packet of energy (or photon) is expelled from a nucleus. There is a tendency to think of radiation as something tangible it has mass and we can discuss the particles involved. Gamma decay, however, deals with energy being expelled, so that the result is a less energetic, more stable nucleus. Beta decay results in a neutron converting into a proton as it emits an electron (beta particle). This changes the nucleus of the original into that of another element with an atomic number one greater than the original. Alpha decay results in the expulsion of a particle of 2 protons and 2 neutrons. This changes the nucleus of the original into that of another element with an atomic number that is two greater than the original. The correct answer is (C).

8 Question No. 8 of Thorium decays through alpha decay to Radium. Which of these represents this radioactive decay process? Thorium has an atomic number of 90 and atomic mass of 230. Question 08 (A) (B) Th Ra + e + ν Th Ra + γ (C) 230 Th 228 Ra+ 1 1 H (D) 230 Th 234 Ra+ 2 4 He (E) 230 Th 22 Ra+ 2 4 He The conversion of a neutron to proton with the emission of an electron and antineutrino is the characteristic of beta decay. This process shows a gamma particle being emitted, but in a gamma decay there is no change in the element, you cannot have two elements with the same atomic mass and atomic number as each other. C Incorrect! Don t get confused, it is not a hydrogen nucleus that is emitted in alpha decay. You are correct that a helium nucleus is emitted when in alpha decay, but what does this do to the atomic number and atomic mass of the daughter nucleus? E. Correct! When Thorium decays through alpha decay, it will emit a helium nucleus which consists of 2 protons and 2 neutrons. The daughter nucleus will have an atomic number that is 2 less than the parent and an atomic mass that is 4 less. When Thorium decays through alpha decay, it will emit a helium nucleus which consists of 2 protons and 2 neutrons. The daughter nucleus will have an atomic number that is 2 less than the parent and an atomic mass that is 4 less Th Ra He The correct answer is (E).

9 Question No. 9 of A friend hypothesizes that after three half lives, about 33% of the original atoms have not decayed. Which of the following best describes a correct reaction on your part? Question 09 (A) You are very intelligent I am glad that you understand this concept so well. (B) I am sorry but you are incorrect, it will all be decayed. (C) I am sorry but you are incorrect because about 13% of the original atoms will not have decayed. (D) I am sorry but you are incorrect because about 25% of the original atoms will not have decayed. (E) I am sorry but you are incorrect because about 87% of the original atoms will not have decayed. It is clear that after two half lives, less than 33% of the original atoms will be left. In fact, that number will be 25%, which is much less and creates an impossibility that 33% would be left after three half lives. Over time, though a sample may approach 100% decay, it will never mathematically be all decayed. Thus this answer is not a possible answer for any question of this type. C. Correct! After two half lives, 25% of the original atoms would remain. A third half life would bring the decay of half of that 25% and leave 12.5% of the original amount. Since this is approximately 13%, this signals a correct answer. After two half lives, 25% of the original atoms remain, but the question asks for the situation after three half lives. The question is talking about how much of the original atoms remain not how many have decay. Half life is the time required for half of the nuclei in a sample of a specific isotope to undergo radioactive decay. So after one half life, 50% of the original is left. After two half lives, half of the 50% or 25% will be left. When working with half life, counting on fingers can sometimes be as valuable as having a calculator. You can use your fingers to keep track of how many times half lives have passed, and mentally or verbally keep cutting the number you are starting with in half. After two half lives, 25% of the original atoms would remain. A third half life would bring the decay of half of that 25% and leave 12.5% of the original amount. Since this is approximately 13%, this signals a correct answer. The correct answer is (C).

10 Question No. 10 of What type of process is represented in each of these cases, which could be used to produce energy? Question 10 (1) (2) (A) (1)Fission, (2) Fission. Both could be used to produce energy. (B) (1) Fusion, (2) Fusion. Neither can be used to produce energy. (C) (1) Fission, (2) Fusion. Both can be used to produce energy. (D) (1) Fusion, (2) Fission. Both can be used to produce energy. (E) (1) Fusion, (2) Fission. Only fusion can be used to produce energy. In one case the nucleus in splitting and in the other two small nuclei are combining to produce a larger nuclei. These are different types of processes Both of these reactions will produce energy. In one case the nucleus in splitting and in the other two small nuclei are combining to produce a larger nuclei. These are different types of processes. Both of these reactions will produce energy. C. Incorrect! Fusion refers to two things coming together. Both of these reactions will produce energy. D. Correct! Fission is the splitting of the nucleus to produce two smaller nuclei, with the release of energy. Fusion is the process of two small nuclei combining to produce a larger nucleus, with the release of energy. Fission is the splitting of the nucleus to produce two smaller nuclei, with the release of energy. Fusion is the process of two small nuclei combining to produce a larger nucleus, with the release of energy. Fission is the splitting of the nucleus to produce two smaller nuclei, with the release of energy. Fusion is the process of two small nuclei combining to produce a larger nucleus, with the release of energy. Fission is the process that is used in nuclear reactors to produce electricity and the first atomic bombs. Fusion is the process that creates the energy in stars such as our sun, its use for power generation is still in the experimentation phase. It is the process that is used in the hydrogen bomb. The correct answer is (D).

MCAT General Chemistry Discrete Question Set 24: Atomic & Nuclear Structure

MCAT General Chemistry Discrete Question Set 24: Atomic & Nuclear Structure Question No. 1 of 10 1. How many protons and neutrons would be found in an atom of Iron with an atomic mass of 56 and an atomic