Every Book Back multiple-choice question from Atomic and Nuclear physics (12th Standard Physics, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus of atomic number Z, then the distance of closest approach of alpha particle to the nucleus is
- A. 14.4 \frac{Z}{V} \text{\AA}Correct
- B. 14.4 \frac{V}{Z} \text{\AA}
- C. 1.44 \frac{Z}{V} \text{\AA}
- D. 1.44 \frac{V}{Z} \text{\AA}
Explanation. At the distance of closest approach, the initial kinetic energy of the alpha particle equals the electrostatic potential energy of the alpha-nucleus system. Solving for distance using the accelerated potential gives this formula in Angstroms.
Q2
In a hydrogen atom, the electron revolving in the fourth orbit, has angular momentum equal to
- A. h
- B. \frac{h}{\pi}
- C. \frac{4h}{\pi}
- D. \frac{2h}{\pi}Correct
Explanation. Bohr's quantization postulate states that angular momentum is an integral multiple of h divided by 2 times pi. For the fourth orbit, substituting four into this relation simplifies to the result.
Q3
Atomic number of H-like atom with ionization potential 122.4 V for n = 1 is
- A. 1
- B. 2
- C. 3Correct
- D. 4
Explanation. The ionization energy of a hydrogen-like atom is proportional to the square of its atomic number. Dividing the given potential by the ground state energy of hydrogen reveals the squared atomic number.
Q4
The ratio between the radius of first three orbits of hydrogen atom is
- A. 1:2:3
- B. 2:4:6
- C. 1:4:9Correct
- D. 1:3:5
Explanation. According to the Bohr model, the radius of an electron's orbit is directly proportional to the square of the principal quantum number. Therefore, the ratio for the first three orbits follows the sequence of squares.
Q5
The charge of cathode rays particle is
- A. positive
- B. negativeCorrect
- C. neutral
- D. not defined
Explanation. Cathode rays consist of electrons, which are subatomic particles carrying a negative electric charge. This was established through experiments involving their deflection in electric and magnetic fields.
Q6
In J.J. Thomson e/m experiment, electrons are accelerated through 2.6 kV enter the region of crossed electric field magnetic field of strength 3.0 × 10^4 Vm–1 and 1.0 × 10^–3 T, respectively, and pass through it and undeflected, then the specific charge is
- A. 1.6 × 10^{10} C kg^{–1}
- B. 1.7 × 10^{11} C kg^{–1}Correct
- C. 1.5 × 10^{11} C kg^{–1}
- D. 1.8 × 10^{11} C kg^{–1}
Explanation. The velocity is found using the ratio of electric to magnetic field strengths. The specific charge is then calculated by relating this velocity and the accelerating potential through the conservation of energy.
Q7
The ratio of the wavelengths radiation emitted for the transition from n =2 to n = 1 in Li++, He+ and H is
- A. 1: 2: 3
- B. 1: 4: 9
- C. 3:2:1
- D. 4: 9: 36Correct
Explanation. The wavelength of emitted radiation is inversely proportional to the square of the atomic number for a given transition. Calculating the inverse squares for lithium, helium, and hydrogen gives the required ratio.
Q8
The electric potential of an electron is given by V = V_0 \ln(r/r_0), where r_0 is a constant. If Bohr atom model is valid, then variation of radius of nth orbit r_n with the principal quantum number n is
- A. r_n \propto \frac{1}{n}
- B. r_n \propto nCorrect
- C. r_n \propto \frac{1}{n^2}
- D. r_n \propto n^2
Explanation. Equating the centripetal force to the force derived from the given logarithmic potential shows that velocity is constant. Applying Bohr's angular momentum quantization then indicates that the radius must vary linearly with n.
Q9
If the nuclear radius of 27Al is 3.6 fermi, the approximate nuclear radius of 64Cu in fermi is
- A. 2.4
- B. 1.2
- C. 4.8Correct
- D. 3.6
Explanation. Nuclear radius is proportional to the cube root of the mass number. By comparing the cube roots of the mass numbers for copper and aluminum, the unknown radius is determined using the given value.
Q10
The nucleus is approximately spherical in shape. Then the surface area of nucleus having mass number A varies as
- A. A^{2/3}Correct
- B. A^{4/3}
- C. A^{1/3}
- D. A^{5/3}
Explanation. The surface area of a sphere depends on the square of its radius. Since the radius of a nucleus is proportional to the cube root of the mass number, the area varies as the mass number raised to two-thirds.
Q11
The mass of a 3Li7 nucleus is 0.042 u less than the sum of the masses of all its nucleons. The average binding energy per nucleon of 3Li7 nucleus is nearly
- A. 46 MeV
- B. 5.6 MeVCorrect
- C. 3.9 MeV
- D. 23 MeV
Explanation. The mass defect is converted to total binding energy using the energy equivalent of one atomic mass unit. Dividing this total energy by the mass number of lithium yields the average binding energy per nucleon.
Q12
M_p denotes the mass of the proton and M_n denotes mass of a neutron. A given nucleus of binding energy B, contains Z protons and N neutrons. The mass M(N,Z) of the nucleus is given by (where c is the speed of light)
- A. M(N,Z) = NM_n + ZM_p - Bc^2
- B. M(N,Z) = NM_n + ZM_p + Bc^2
- C. M(N,Z) = NM_n + ZM_p - B/c^2Correct
- D. M(N,Z) = NM_n + ZM_p + B/c^2
Explanation. Binding energy is defined as the difference between the total mass of individual nucleons and the actual mass of the nucleus, multiplied by c squared. Rearranging this definition allows the calculation of the nuclear mass.
Q13
A radioactive nucleus (initial mass number A and atomic number Z) emits two alpha-particles and 2 positrons. The ratio of number of neutrons to that of proton in the final nucleus will be
- A. \frac{A-Z-4}{Z-2}
- B. \frac{A-Z-2}{Z-6}Correct
- C. \frac{A-Z-4}{Z-6}
- D. \frac{A-Z-12}{Z-4}
Explanation. Each alpha decay reduces the mass number by four and atomic number by two, while positron emission reduces only the atomic number by one. Tracking these changes for all emissions allows calculating the final neutron and proton counts.
Q14
The half-life period of a radioactive element A is same as the mean life time of another radioactive element B. Initially both have the same number of atoms. Then
- A. A and B have the same decay rate initially
- B. A and B decay at the same rate always
- C. B will decay at faster rate than ACorrect
- D. A will decay at faster rate than B
Explanation. The decay constant is inversely proportional to the mean life and the half-life. Since the half-life of A equals the mean life of B, the decay constant of B is greater, meaning B decays more rapidly.
Q15
A radioactive element has N_0 number of nuclei at t=0. The number of nuclei remaining after half of a half-life (that is, at time t = \frac{1}{2} T_{1/2}) is
- A. \frac{N_0}{2}
- B. \frac{N_0}{\sqrt{2}}Correct
- C. \frac{N_0}{4}
- D. \frac{N_0}{8}
Explanation. Using the law of radioactive decay, the remaining fraction of nuclei is calculated by substituting the specific time interval. For half of a half-life, the remaining amount involves an exponential factor equal to the inverse square root of two.
