Nuclei-Quiz-23

 

NUCLEI QUIZ-23

Dear Readers,

JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get a detailed list of all topics that are important in cracking the entrance examination. JEE Advanced syllabus for Physics has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with day-to-day experiences. In comparison to the other two subjects, the syllabus of JEE Advanced for physics is developed in such a way so as to test the deep understanding and application of concepts.

Q1. The β-decay process, discovered around 1900, is basically the decay of a neutron (n), in the laboratory, a proton (p) and an electron (e^-) are observed as the decay products of the neutron therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e.n→p+e^-+v ̅_e, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino (v ̅_e) to be massless and possessing negligible energy, and neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8×10^6 eV. The kinetic energy carried by the proton is only the recoil energy.

   What is the maximum energy of the anti-neutrino

•   Zero
•   Much less than 0.8×10^6 eV
•   Nearly 0.8×10^6 eV
•   Much larger than 0.8×10^6 eV

Solution

Q2. When subatomic particles undergo reactions, energy is conserved, but mass is not necessarily conserved. However, a particle’s mass ‘contributes’ to its total energy, in accordance with Einstein’s famous equation, E=mc^2 In this equation, E denotes the energy a particle carries because of its mass. The particle can also have additional energy due to its motion and its interaction with order particles Consider a neutron at rest, and well separated from other particles. It decays into a proton, an electron, and an undetected third particle: Neutron→proton+electron+??? The table below summarizes some data from a single neutron decay. An MeV (mega electron volt) is a unit energy. Column 2 shows the rest mass of the particle times the speed of light squared

Assuming the table contains no major errors, what can we conclude about the (mass×c^2) of the undetected third particle?

•   It is 0.79 MeV
•   It is 0.39 MeV
•   It is less than or equal to 0.79 MeV; but we cannot be more precise
•   It is less than or equal to 0.39 MeV; but we cannot be more precise

Solution

(d) According to the passage, subatomic reactions do not conserve mass. So, we cannot find the third particle’s mass by setting m_neutron equal to m_proton+m_electron+m_(third particle). By constrast, the total energy in this case, the sum of ‘mass energy’ and kinetic energy, is conserved. If E denotes total energy, then E_neutron=E_proton+E_electron+E_(third particle) The neutron has energy 949.97 MeV. The proton has energy 939.67 MeV+0.01 MeV=939.69 MeV. The electron has energy 0.51 MeV=0.39 MeV=0.90 MeV. Therefore, the third particle has energy E_(third particle)=E_neutron-E_proton-Electron We just found the third particle’s total energy, the sum of its mass energy and kinetic energy. Without more information, we cannot figure out how much of that energy is mass energy

Q3.The compound unstable nucleus _92^236 U often decays in accordance with the following reaction: _92^236 U→ _54^140 Xe+ _38^94 Sr+ other particles During the reaction, the uranium nucleus ‘fission’ (splits) into the two smaller nuclei. The reaction is energetically favorable because the small nuclei have higher nuclear binding energy per nucleon (although the lighter nuclei have lower total nuclear binding energies, because they contain fewer nucleons) Inside a nucleus, the nucleons (protons and neutrons) attract each other with a ‘strong nuclear’ force. All nucleons exert approximately the same strong nuclear force on each other. This force holds the nucleus together. Importantly, the strong nuclear force becomes important only when the protons and neutrons are very close together at intranuclear distances In the nuclear reaction presented above, the ‘other particles’ might be

•   An alpha particle, which consists of two protons and neutrons
•   Two protons
•   One proton and one neutron
•   Two neutrons

Solution (d) Nuclear reactions conserve total charge and also conserve the total approximate mass (as measured by the atomic mass number). Therefore, since the uranium, xenon, and strontium nuclei have atomic masses 236, 140 and 94, the ‘other particles’ must have total atomic mass A such that 236=140+94+A So, A=2. The other particles are two nucleons. This narrows down the answer to options (b), (c) and (d). For nuclei, the atomic number -i.e., the number of protons-tells us the charge. So, the other particles must have total charge Z such that 92=54+38+Z or Z=0 In summary, the other particles have total atomic mass 2 and total charge 0. Only option (d) fits this description

Q4. A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in ‘head-on’ to a particular lead nucleus and stops 6.50×10^(-14) m away from the center of the nucleus.(The point is well outside the nucleus.) Assuming that the lead nucleus, which has 82 protons, remains at rest. The mass of alpha particle is 6.64×10^(-27)kg 266.

 Calculate the electrostatic potential energy at the instant when the alpha particle stops?

•   36.3 MeV
•   45.0 MeV
•   3.63 MeV
•   40.0 MeV

Solution

Q5. A nucleus, kept at rest in free space, break up into two smaller nuclei of masses m and 2m. Total energy generated in this fission is E. The bigger part is radioactive, emits five gamma ray photons in the direction opposite to its velocity, and finally comes to rest. Now, answer the following questions: (given:h=6.6×10^(-34) J s, m=1.00×10^(-26) kg,E=3.63×10^(-8) mc^2,c=3×10^8 ms^(-1) Fractional loss of mass in the fission is

•   1.21×10^(-8)
•   2.56×10^(-8)
•   1.73×10^(-8)  
•   3.52×10^(-8)

Solution

(a) Use conservation of energy and momentum Momentum of a photon =h/λ  

Q6.The results of activity measurements on a radioactive sample are given in the table below.

The half-life of the radioactive nuclei is nearly (ln⁡〖2=0.693,ln⁡〖3=1.0986)〗 〗

•   2.5 h
•   7 h
• 5 h
•   1.2 h

Solution

Q7.Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, _1^2 H, known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D reaction is _1^2 H+ _1^2 H→ _2^3 He+n+ energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of _1^2 H nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time t_0 before the particles fly away from the core. If n is the density (number/volume) of deuterons, the product nt_0 is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5×10^14 s/cm^3. It may be helpful to use the following : Boltzmann constant k=8.6×10^(-5) eV/K;e^2=1.44×10^(-9) eVm 269. In the core of nuclear fusion reactor, the gas becomes plasma because of

•  Strong nuclear force acting between the deuterons
•   Coulomb force acting between the deuterons
•   Coulomb force acting between deuterons-electron pairs
•   The high temperature maintained inside the reactor core

Solution

(d) The high temperature maintained inside the reactor core

Q8. When a particle is restricted to move along x-axis between x=0 and x=a, where a is nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x=0 and x=a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as E=p^2/2m. Thus, the energy of the particle can be denoted by a quantum number n' taking values 1,2,3,……(n=1, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving in the line x=0 to x=a. Take h=6.6×10^(-34) J s and e=1.6×10^(-19) C The allowed energy for the particle for a particular value of n is proportional to

•   a^(-2)
•   a^(-3/2)
•   a^(-1)
•   a^2

Solution

Q9. In a mixture of H-He^+ gas (He^+ is singly ionized He atom), H atoms and He^+ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He^+ ions (by collisions). Assume that the Bohr model of atom is exactly valid . The quantum number n of the state finally populated in He^+ ions is

•   2
•   3
•   4
•   5

Solution

Q10. The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr’s quantization condition 272. A diatomic molecule has moment of inertia I. By Bohr’s quantization condition its rotational energy in the n^th level (n=0 is not allowed) is

•  1/n^2 (h^2/(8π^2 I))
•   1/n (h^2/(8π^2 I))
• n(h^2/(8π^2 I))
• n^2 (h^2/(8π^2 I))
  
  

Solution