Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
The binding energy per nucleon of $^{209}_{83}Bi$ is ________ MeV. [Take $m(^{209}_{83}Bi) = 208.980388$ u, $m_p = 1.007825$ u, $m_n = 1.008665$ u, $1$ u $= 931$ MeV/c$^2$]
(1) 7.48
(2) 7.84
(3) 8.79
(4) 6.94
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Answer: Option (2)
Key idea: Binding Energy Per Nucleon and Mass Defect
Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the $2^{nd}$ and $4^{th}$ orbits of hydrogen atom ________.
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Answer: 32
Key idea: Magnetic field due to electron motion in Bohr orbits
The electron in orbit acts as a current loop with current $I_n = \frac{e}{T_n}$, where $T_n$ is the period of revolution
From Bohr's model: radius $r_n = n^2a_0$ and velocity $v_n = \frac{v_1}{n}$
Period: $T_n = \frac{2\pi r_n}{v_n} = \frac{2\pi n^3 a_0}{v_1}$, so current $I_n = \frac{ev_1}{2\pi n^3 a_0}$
Magnetic field at center: $B_n = \frac{\mu_0 I_n}{2r_n} = \frac{\mu_0 ev_1}{4\pi n^5 a_0^2}$
Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass numbers 3, 4 and 10 are 5.6 MeV, 7.4 MeV and 6.1 MeV, respectively, then in the process, $\Delta Mc^2 = $ ______ MeV.
(1) $6.9$
(2) $7.9$
(3) $2.2$
(4) $4.3$
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Answer: Option (3)
Key idea: Nuclear Binding Energy and Mass-Energy Equivalence
Write the reaction: $2(A=3) + (A=4) \to (A=10)$
Calculate total binding energies: BE = (mass number) × (BE per nucleon)
In Rutherford's alpha-particle scattering experiment, only a few alpha particles rebound back because
A. The size of gold nucleus is very small as compared to the size of gold atom.
B. Alpha particle and gold nucleus have equal charge.
C. The impact parameter is minimum for a few alpha particles.
D. A few alpha particles have very high kinetic energy.
E. Only a few alpha particles undergo head-on collision with the nuclei.
Choose the correct answer from the options given below:
(1) A, B Only
(2) B, E Only
(3) C, D Only
(4) A, C, E Only
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Answer: Option (4)
Key idea: Rutherford's Alpha Particle Scattering Experiment - Conditions for Large Angle Scattering
In Rutherford's experiment, alpha particles are scattered by the gold nucleus through Coulomb repulsion
Statement A is correct: The nucleus ($\sim 10^{-15}$ m) is much smaller than the atom ($\sim 10^{-10}$ m), so probability of close approach is very small
Statement B is incorrect: Alpha particle has charge $+2e$ while gold nucleus has charge $+79e$ (not equal)
Statement C is correct: Large angle scattering (rebounding) requires very small impact parameter $b$, which only a few particles have
Statement D is incorrect: All alpha particles from the source have approximately the same kinetic energy
Statement E is correct: Head-on or nearly head-on collisions (equivalent to small impact parameter) are rare due to small nuclear size
Therefore, statements A, C, and E correctly explain why only few alpha particles rebound
In the hydrogen atom, the electron makes a transition from the higher orbit $(i)$ to a lower orbit $(f)$. The ratio of the radius of the orbits in given by $r_i : r_f = 16 : 4$. The wavelength of photon emitted due to this transition is ______ nm. (Given Rydberg constant = $1.0973 \times 10^7$ /m)
(1) 121
(2) 242
(3) 486
(4) 974
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Answer: Option (3)
Key idea: Bohr Model: Orbital radius and Rydberg formula for spectral transitions
Given $r_i : r_f = 16 : 4 = 4 : 1$. Using Bohr's model, $r_n \propto n^2$, so $\frac{r_i}{r_f} = \frac{n_i^2}{n_f^2} = 4$
This gives $\frac{n_i}{n_f} = 2$, meaning $n_i = 2n_f$. For $n_f = 2$, we get $n_i = 4$ (transition from $n=4$ to $n=2$)
Apply Rydberg formula: $\frac{1}{\lambda} = R \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$
Assuming the experimental mass of $^{12}_6$C as 12 u, the mass defect of $^{12}_6$C atom is _______MeV/c$^2$.
(Mass of proton = 1.00727 u, mass of neutron = 1.00866 u, 1 u = 931.5 MeV/c$^2$ and c is the speed of the light in vacuum).
(1) 127.5
(2) 89.03
(3) 272.0
(4) 92.0
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Answer: Option (2)
Key idea: Mass Defect and Nuclear Binding Energy
Identify that $^{12}_6$C has 6 protons and 6 neutrons (A - Z = 12 - 6 = 6)
Calculate total mass of separated nucleons: $6 \times 1.00727 + 6 \times 1.00866 = 12.09558$ u
Find mass defect: $\Delta m = 12.09558 - 12 = 0.09558$ u
Convert to MeV/c²: $0.09558 \times 931.5 = 89.03$ MeV/c²
The ratio of momentum of the photons of the $1^{st}$ and $2^{nd}$ line of Balmer series of Hydrogen atoms is $\alpha/\beta$. The possible values of $\alpha$ and $\beta$ are:-
(1) 27 and 20
(2) 3 and 16
(3) 5 and 36
(4) 20 and 27
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Answer: Option (4)
Key idea: Balmer Series and Photon Momentum
Identify transitions: 1st Balmer line is $n=3 \to n=2$, 2nd line is $n=4 \to n=2$
Calculate energy for 1st line: $E_1 \propto \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{5}{36}$
Calculate energy for 2nd line: $E_2 \propto \frac{1}{2^2} - \frac{1}{4^2} = \frac{1}{4} - \frac{1}{16} = \frac{3}{16}$
Since photon momentum $p = \frac{E}{c}$, the ratio $\frac{p_1}{p_2} = \frac{E_1}{E_2}$
The energy released if hydrogen atoms are combined to form $\frac{4}{2}$He is _______ MeV.
(Take binding energies per nucleon of $\frac{2}{1}$H and $\frac{4}{2}$He as 1.1 MeV and 7.2 MeV, respectively)
(1) 6.1
(2) 24.4
(3) 26.6
(4) 5
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Answer: Option (2)
Key idea: Nuclear Binding Energy and Mass-Energy Equivalence
Identify the reaction: $2(^2_1H) \to ^4_2He$
Calculate total BE of reactants: $2 \times 2 \times 1.1 = 4.4$ MeV
Calculate total BE of product: $4 \times 7.2 = 28.8$ MeV
Energy released $= BE_{product} - BE_{reactants} = 28.8 - 4.4 = 24.4$ MeV
Two radioactive substances A and B of mass numbers 200 and 212 respectively, shows spontaneous $\alpha$-decay with same $Q$ value of 1 MeV. The ratio of energies of $\alpha$-rays produced by A and B is _______.
(1) $\frac{2548}{2650}$
(2)
(3)
(4)
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Answer: Option (1)
Key idea: Energy distribution in alpha decay using conservation of momentum and energy
In α-decay, the Q-value is shared between the α-particle and recoil nucleus according to momentum conservation
Using $p_\alpha = p_{recoil}$ and energy conservation, derive: $K_\alpha = Q \times \frac{m_{daughter}}{m_{daughter} + m_\alpha}$
