MCQ Question for Class 12 Physics Chapter 12 Atoms

MCQs MCQs Class 12

Refer to MCQ Class 12 Atoms provided below which is an important chapter in Class 12 Physics. Students should go through the MCQs questions for Class 12 Physics Chapter 12 Atoms with answers given below so that they are able to understand the complete topic properly. It’s important to understand the entire chapter by reading Class 12 Physics Notes also. Also, refer to MCQ Questions for Class 12 Physics for all chapters.

MCQ on Atoms Class 12 Physics PDF with Answers

All multiple choice questions with solutions provided below have been developed based on the latest syllabus and examination pattern issued for class 12 by CBSE and NCERT. As Atoms is a very important and scoring chapter in Physics Class 12, therefore, the students should carefully learn the questions and answers given below which will help them to get better scores in upcoming examinations for class 12th.

Question. When an a-particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as
(a) 1/m2
(b) m
(c) 1/m
(d) 1/√m

Answer

C

Question. An alpha nucleus of energy 1/2 mv2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
(a) 1/Ze
(b) v2
(c) 1/m
(d) 1/v4

Answer

C

Question. In a Rutherford scattering experiment when a projectile of charge z1 and mass M1 approaches a target nucleus of charge z2 and mass M2, the distance of closest approach is r0. The energy of the projectile is
(a) directly proportional to z1z2
(b) inversely proportional to z1
(c) directly proportional to mass M1
(d) directly proportional to M1 × M2

Answer

A

Question. The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is
(a) 1
(b) 4
(c) 0.5
(d) 2

Answer

B

Question. Given the value of Rydberg constant is 107 m–1, the wave number of the last line of the Balmer series in hydrogen spectrum will be
(a) 0.25 × 107 m–1
(b) 2.5 × 107 m–1
(c) 0.025 × 104 m–1
(d) 0.5 × 107 m–1

Answer

A

Question. Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is
(a) 7/29
(b) 9/31
(c) 5/27
(d) 5/23

Answer

C

Question. The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion is
(a) 3
(b) 4
(c) 1
(d) 2 

Answer

D

Question. Which source is associated with a line emission spectrum?
(a) Electric fire
(b) Neon street sign
(c) Red traffic light
(d) Sun

Answer

B

Question. For which one of the following, Bohr model is not valid?
(a) Hydrogen atom
(b) Singly ionised helium atom (He+)
(c) Deuteron atom
(d) Singly ionised neon atom (Ne+) 

Answer

D

Question. The total energy of an electron in an atom in an orbit is – 3.4 eV. Its kinetic and potential energies are, respectively
(a) 3.4 eV, 3.4 eV
(b) – 3.4 eV, – 3.4 eV
(c) – 3.4 eV, – 6.8 eV
(d) 3.4 eV, – 6.8 eV

Answer

D

Question. The radius of the first permitted Bohr orbit for the electron, in a hydrogen atom equals 0.51 Å and its ground state energy equals –13.6 eV. If the electron in the hydrogen atom is replaced by muon (m–) [charge same as electron and mass 207 me], the first Bohr radius and ground state energy will be
(a) 0.53 × 10–13 m, –3.6 eV
(b) 25.6 × 10–13 m, –2.8 eV
(c) 2.56 × 10–13 m, –2.8 keV
(d) 2.56 × 10–13 m, –13.6 eV 

Answer

C

Question. The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is
(a) 1 : 1
(b) 1 : –1
(c) 2 : –1
(d) 1 : –2 

Answer

B

Question. Consider 3rd orbit of He+ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K = 9 × 109 constant, Z = 2 and h (Planck’s constant) = 6.6 × 10–34 J s]
(a) 0.73 × 106 m/s
(b) 3.0 × 108 m/s
(c) 2.92 × 106 m/s
(d) 1.46 × 106 m/s

Answer

D

Question. An electron in hydrogen atom makes a transition n1 → n2 where n1 and n2 are principal quantum numbers of the two states. Assuming Bohr’s model to be valid, the time period of the electron in the initial state is eight times that in the final state. The possible values of n1 and n2 are
(a) n1 = 6 and n2 = 2
(b) n1 = 8 and n2 = 1
(c) n1 = 8 and n2 = 2
(d) n1 = 4 and n2 = 2

Answer

D

Question. Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material.
The stopping potential is measured to be 3.57 V. The threshold frequency of the material is
(a) 4 × 1015 Hz
(b) 5 × 1015 Hz
(c) 1.6 × 1015 Hz
(d) 2.5 × 1015 Hz 

Answer

C

Question. An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, then the value of n is
(a) 2
(b) 3
(c) 4
(d) 5 

Answer

C

Question. Out of the following which one is not a possible energy for a photon to be emitted by hydrogen atom according to Bohr’s atomic model?
(a) 0.65 eV
(b) 1.9 eV
(c) 11.1 eV
(d) 13.6 eV 

Answer

C

Question. The ionization energy of Li++ is equal to
(a) 9hcR
(b) 6hcR
(c) 2hcR
(d) hcR.

Answer

A

Question. The Rutherford a-particle experiment shows that most of the a-particles pass through almost unscattered while some are scattered through large angles. What information does it give about the structure of the atom?
(a) Atom is hollow.
(b) The whole mass of the atom is concentrated in a small centre called nucleus
(c) Nucleus is positively charged
(d) All the above

Answer

D

Question. The energy of a hydrogen atom in the ground state is –13.6 eV. The energy of a He+ ion in the first excited state will be
(a) –13.6 eV
(b) –27.2 eV
(c) –54.4 eV
(d) –6.8 eV 

Answer

A

Question. The electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping potential is estimated to be (the energy of the electron in nth state En = −13.6/ n2. eV )
(a) 5.1 V
(b) 12.1 V
(c) 17.2 V
(d) 7 V 

Answer

D

Question. The ground state energy of hydrogen atom is –13.6 eV. When its electron is in the first excited state, its excitation energy is
(a) 10.2 eV
(b) 0
(c) 3.4 eV
(d) 6.8 eV 

Answer

A

Question. The total energy of electron in the ground state of hydrogen atom is –13.6 eV. The kinetic energy of an electron in the first excited state is
(a) 6.8 eV
(b) 13.6 eV
(c) 1.7 eV
(d) 3.4 eV 

Answer

D

Question. The total energy of an electron in the first excited state of hydrogen atom is about –3.4 eV. Its kinetic energy in this state is
(a) 3.4 eV
(b) 6.8 eV
(c) –3.4 eV
(d) –6.8 eV

Answer

A

Question. The Bohr model of atoms
(a) Assumes that the angular momentum of electrons is quantized.
(b) Uses Einstein’s photoelectric equation.
(c) Predicts continuous emission spectra for atoms.
(d) Predicts the same emission spectra for all types of atoms. 

Answer

A

Question. In which of the following systems will the radius of the first orbit (n = 1) be minimum?
(a) doubly ionized lithium
(b) singly ionized helium
(c) deuterium atom
(d) hydrogen atom

Answer

A

Question. The third line of Balmer series of an ion equivalnet to hydrogen atom has wavelength of 108.5 nm. The ground state energy of an electron of this ion will be
(a) 3.4 eV
(b) 13.6 eV
(c) 54.4 eV
(d) 122.4 eV

Answer

C

Question. The first line of Balmer series has wavelength 6563 Å. What will be the wavelength of the first member of Lyman series
(a) 1215.4 Å
(b) 2500 Å
(c) 7500 Å
(d) 600 Å

Answer

A

Question. The energy of hydrogen atom in nth orbit is En then the energy in nth orbit of singly ionised helium atom will be
(a) 4En
(b) En/4
(c) 2En
(d) En/2 

Answer

A

Question. The life span of atomic hydrogen is
(a) fraction of one second
(b) one year
(c) one hour
(d) one day

Answer

A

Question. The energy of the ground electronic state of hydrogen atom is –13.6 eV. The energy of the first excited state will be
(a) –27.2 eV
(b) –52.4 eV
(c) –3.4 eV
(d) –6.8 eV

Answer

C

Question. When hydrogen atom is in its first excited level, its radius is ………. of the Bohr radius.
(a) twice
(b) 4 times
(c) same
(d) half 

Answer

B

Question. According to Bohr’s principle, the relation between principal quantum number (n) and radius of orbit (r) is
(a) r ∝ 1/n
(b) r ∝ 1/n2
(c) r ∝ n
(d) r ∝ n2 

Answer

D

Question. In a Rutherford experiment, the number of particles scattered at 90° angle are 28 per minute then number of scattered particles at an angle 60° and 120° will be
(a) 117 per minute, 25 per minute
(b) 50 per minute, 12.5 per minute
(c) 100 per minute, 200 per minute
(d) 112 per minute, 12.4 per minute

Answer

D

Question. In Hydrogen spectrum, the wavelength of Ha line is 656 nm, whereas in the spectrum of a distant galaxy, Ha line wavelength is 706 nm. Estimated speed of the galaxy with respect to earth is
(a) 2 × 108 m/s
(b) 2 × 107m/s
(c) 2 × 106 m/s
(d) 2 × 105 m/s

Answer

B

Question. When a hydrogen atom is raised from the ground state to an excited state,
(a) both K.E. and P.E. increase
(b) both K.E. and P.E. decrease
(c) the P.E. decreases and K.E. increases
(d) the P.E. increases and K.E. decreases. 

Answer

D

Question. In terms of Bohr radius a0, the radius of the second Bohr orbit of a hydrogen atom is given by
(a) 4a0
(b) 8a0
(c) 2a0
(d) 2a0 

Answer

A

Question. The ionization energy of hydrogen atom is 13.6 eV.
Following Bohr’s theory, the energy corresponding to a transition between 3rd and 4th orbit is
(a) 3.40 eV
(b) 1.51 eV
(c) 0.85 eV
(d) 0.66 eV 

Answer

D

Question. The ground state energy of H-atom is –13.6 eV. The energy needed to ionize H-atom from its second excited state
(a) 1.51 eV
(b) 3.4 eV
(c) 13.6 eV
(d) none of these 

Answer

A

Question. To explain his theory, Bohr used
(a) conservation of linear momentum
(b) quantisation of angular momentum
(c) conservation of quantum frequency
(d) none of these

Answer

B

Question. Excitation energy of a hydrogen like ion in its excitation state is 40.8 eV. Energy needed to remove the electron from the ion in ground state is
(a) 54.4 eV
(b) 13.6 eV
(c) 40.8 eV
(d) 27.2 eV

Answer

A

Question. The ionisation potential of H-atom is 13.6 V. When it is excited from ground state by monochromatic radiations of 970.6 Å, the number of emission lines will be (according to Bohr’s theory)
(a) 10
(b) 8
(c) 6
(d) 4

Answer

C

Question. The ionisation energy of hydrogen atom is 13.6 eV,the ionisation energy of a singly ionised helium atom would be
(a) 13.6 eV
(b) 27.2 eV
(c) 6.8 eV
(d) 54.4 eV

Answer

D

Question. If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength l. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be
(a) 16/25λ
(b) 9/16λ
(c) 20/7λ
(d) 20/13λ

Answer

C

Question. Hydrogen atom in ground state is excited by a monochromatic radiation of l = 975 Å. Number of spectral lines in the resulting spectrum emitted will be
(a) 3
(b) 2
(c) 6
(d) 10 

Answer

C

Question. Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths l1 : l2 emitted in the two cases is
(a) 7/5
(b) 27/20
(c) 27/5
(d) 20/7 

Answer

D

Question. An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be
(a) 24hR/25m
(b) 25hR/24m
(c) 25m/24hR
(d) 24m/25hR
(m is the mass of the electron, R Rydberg constant and h Planck’s constant)

Answer

A

Question. The transition from the state n = 3 to n = 1 in a hydrogen like atom results in ultraviolet radiation.
Infrared radiation will be obtained in the transition from
(a) 2 → 1
(b) 3 → 2
(c) 4 → 2
(d) 4 → 3 

Answer

D

Question. A hydrogen atom in its ground state absorbs 10.2 eV of energy. The orbital angular momentum is increased by
(a) 1.05 × 10–34 J-s
(b) 3.16 × 10–34 J-s
(c) 2.11 × 10–34 J-s
(d) 4.22 × 10–34 J-s

Answer

A

Question. Taking Rydberg’s constant RH = 1.097 × 107m, first and second wavelength of Balmer series in hydrogen spectrum is
(a) 2000 Å, 3000 Å
(b) 1575 Å, 2960 Å
(c) 6529 Å, 4280 Å
(d) 6552 Å, 4863 Å

Answer

B

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