CBSE Board Exam 2020: Important MCQs (with Answers) for Class 12 Physics - Chapter 12 - Atoms; Also useful for JEE Main, UPSEE, WBJEE & Other Engineering Entrance Exams
Check important MCQs (with Answers) for Class 12 Physics Board Exam 2020 (Chapter 12 - Atoms). These MCQs are very useful for other competitive exams like UPSEE 2020, JEE Main 2020, WBJEE 2020 etc.
Learn important MCQs (with solutions) for CBSE Class 12th Physics Board Exam 2020 (Chapter 12 - Atoms). Here you will also get important links to access some important articles for the preparation of CBSE 12th board exams 2020. Students preparing for CBSE Class 12th Physics Board Exam 2020 usually ask about important MCQs (Multiple Choice Questions) & here we have provided important questions (with solution), based on Chapter 12 (Atoms) of Class 12th Physics NCERT textbook.
Important MCQs for CBSE Class 12 Physics Board Exam 2020 (Chapter 12 - Atoms):
Q1. The radius of innermost electron orbit of a hydrogen atom is 5.310-11m. What is the radius of orbit in second excited state?
(a) 5.1 × 10‒5 m
(b) 1.44 × 10‒5 m
(c) 3.1 × 10‒8 m
(d) 4.77 × 10‒10 m
Q2. The number of alpha particles scattered at 60o is 100 per minute in an alpha particle scattering experiment. The number of alpha particles scattered per minute at 90o are:
Q3. The ionization energy of electron in hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of wavelengths Maximum wave- length of emitted radiation corresponds to the transition between
(a) n = 4 to n = 3 states
(b) n = 3 to n = 2 states
(c) n = 3 to n = 31 states
(d) n = 2 to n = 1 states
Maximum wavelength corresponds to minimum energy. The required transition must be from n = 4 to n = 3 states.
Q4. The energy level for an electron in a certain atom is shown in the figure given below. Which of the transition shown in the figure represents the emission of a photon with maximum energy?
The photon emitted is with maximum energy in case of transition III.
E = E2 – E1 = (-13.6/22) – (-13.6/12) = 10.2eV
Q5. The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultra = violet radiation. Infrared radiation will be obtained in the transition from.
(a) 3 – 2
(b) 2 – 1
(c) 4 – 2
(d) 5 – 4
Infra red corresponds to the least value of [1/n12 – 1/n22] Here, it is 5 – 4.
Q6. The total energy of electron in the ground state of hydrogen atom is – 13.6 e V. What is the K.E. of this electron in first excited state?
(a) 13.6 eV
(b) 6.8 eV
(c) 1.7 eV
(d) 3.4 eV
Total energy = P.E. + K.E.
‒13.6 eV = ‒ 2 KE + KE = ‒ K. E
∴ KE = + 13.6 eV
In first excited state, n = 2
∴ K.E. = (13.6/22) ev = 3.4 eV.
Q7. The energy of an electron in nth orbit of hydrogen atom is – 13.6/n2 eV. Energy required to excite thee electron from the first orbit to the third orbit is
(a) 10.2 J
(b) 12.09 J
(c) 12.09 eV
(d) 13.6 eV
E = E3 – E1 = (-13.6/32) – (-13.6/12) = 12.09eV
Q8. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because
(a) of the electrons not being subject to a central force
(b) of the electrons colliding with each other
(c) of screening effects
(d) the force between the nucleus and an electron will no longer be given by Coulomb’s law
The simple Bohr model cannot be directly applied to calculate energy levels of an atom with many electrons because electrons experience force due to electrons in neighbouring electrons in different shells.
Q9. The ground state energy of hydrogen atom is – 13.6 eV. The kinetic and potential energies of the electron in this state are
(a) 13.6 eV and 27.2 eV respectively
(b) ‒ 13.6 eV and ‒ 27.2 eV respectively
(c) 13.6 eV and ‒ 27.2 eV respectively
(d) 27.2 eV and 13.6 eV
Total energy, E = ‒13.6 eV
Kinetic energy, = ‒E = ‒ (‒13.6 eV) = 13.6 eV
Potential energy = ‒ 2 × Kinetic energy = ‒2 13.6 = ‒ 27.2 eV
Q10. Which level of the doubly ionized lithium has the same energy as the ground state energy of the hydrogen atom?