JEE Physics Atoms MCQs Set 21

Physics Objective Questions and Answers: Atoms

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Multiple Choice Questions

Question 1: The acceleration of electron in the first orbit of hydrogen atom is
(a) \( \dfrac{4\pi^2m}{h^3} \)
(b) \( \dfrac{h^2}{4\pi^2mr} \)
(c) \( \dfrac{h^2}{4\pi^2m^2r^3} \)
(d) \( \dfrac{m^2h^2}{4\pi^2r^3} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{h^2}{4\pi^2m^2r^3} \)

Explanation:
1. Bohr's condition: \( mvr=\dfrac{h}{2\pi}\Rightarrow v=\dfrac{h}{2\pi mr} \). Then \( a=\dfrac{v^2}{r}=\dfrac{h^2}{4\pi^2m^2r^3} \).

Teacher's Note:
1. Centripetal acceleration is \( v^2/r \).

Question 2: An electron collides with a hydrogen atom in its ground state and excites it to \( n=3 \). The energy given to hydrogen atom in this inelastic collision is (neglect the recoiling of hydrogen atom)
(a) 10.2 eV
(b) 12.1 eV
(c) 12.5 eV
(d) none of these
Show Answer & Explanation

Answer: (b) 12.1 eV

Explanation:
1. The atom gains \( E_3-E_1=-1.51-(-13.6)=12.1 \) eV.

Teacher's Note:
1. Energy absorbed equals the level gap.

Question 3: An electron of charge e moves with a constant speed v along a circle of radius r, its magnetic moment will be
(a) \( evr \)
(b) \( evr/2 \)
(c) \( \pi r^2ev \)
(d) \( 2\pi rev \)
Show Answer & Explanation

Answer: (b) \( evr/2 \)

Explanation:
1. \( I=\dfrac{ev}{2\pi r} \), \( M=IA=\dfrac{ev}{2\pi r}\pi r^2=\dfrac{evr}{2} \).

Teacher's Note:
1. Magnetic moment is current times area.

Question 4: The orbital frequency of an electron in the hydrogen atom is proportional to
(a) \( n^3 \)
(b) \( n^{-3} \)
(c) \( n \)
(d) \( n^0 \)
Show Answer & Explanation

Answer: (b) \( n^{-3} \)

Explanation:
1. \( f=\dfrac{v}{2\pi r} \) with \( v\propto\dfrac1n \) and \( r\propto n^2 \), so \( f\propto n^{-3} \).

Teacher's Note:
1. Combine the n-dependence of v and r.

Question 5: The product of linear momentum and angular momentum of an electron of the hydrogen atom is proportional to \( n^x \), where x is
(a) 0
(b) 1
(c) \( -2 \)
(d) 2
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. \( p\,L=(mv)(mvr)=m^2v^2r \). With \( v\propto n^{-1} \), \( r\propto n^2 \): \( v^2r\propto n^0 \), so \( x=0 \).

Teacher's Note:
1. Use \( v\propto1/n \) and \( r\propto n^2 \).

Question 6: Which of the following transitions in Balmer series for hydrogen will have longest wavelength?
(a) \( n=2 \) to \( n=1 \)
(b) \( n=6 \) to \( n=1 \)
(c) \( n=3 \) to \( n=2 \)
(d) \( n=6 \) to \( n=2 \)
Show Answer & Explanation

Answer: (c) \( n=3 \) to \( n=2 \)

Explanation:
1. Balmer lines end on \( n=2 \); the longest wavelength is the smallest energy gap, \( 3\to2 \).

Teacher's Note:
1. Options (a) and (b) are not Balmer transitions.

Question 7: An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transitions. The ratio of the velocities of the electron in these two orbits is
(a) \( \tfrac12 \)
(b) \( \tfrac21 \)
(c) \( \tfrac54 \)
(d) \( \tfrac34 \)
Show Answer & Explanation

Answer: (d) \( \tfrac34 \)

Explanation:
1. \( \dfrac{n(n-1)}{2}=6\Rightarrow n=4 \); \( =3\Rightarrow n=3 \). Since \( v\propto\dfrac1n \), \( \dfrac{v_4}{v_3}=\dfrac34 \).

Teacher's Note:
1. Count the transitions from the top orbit.

Question 8: \( \nu_1 \) is the frequency of the series limit of Lyman series, \( \nu_2 \) is the frequency of the first line of Lyman series and \( \nu_3 \) is the frequency of the series limit of the Balmer series. Then,
(a) \( \nu_1-\nu_2=\nu_3 \)
(b) \( \nu_1=\nu_2-\nu_3 \)
(c) \( \dfrac1{\nu_2}=\dfrac1{\nu_1}+\dfrac1{\nu_3} \)
(d) \( \dfrac1{\nu_1}=\dfrac1{\nu_2}+\dfrac1{\nu_3} \)
Show Answer & Explanation

Answer: (a) \( \nu_1-\nu_2=\nu_3 \)

Explanation:
1. \( \nu_1=Rc \), \( \nu_2=\tfrac34Rc \), \( \nu_3=\tfrac14Rc \). So \( \nu_1-\nu_2=\tfrac14Rc=\nu_3 \).

Teacher's Note:
1. Write each in terms of \( Rc \).

Question 9: The spectrum of an oil flame is an example for
(a) line emission spectrum
(b) continuous emission spectrum
(c) line absorption spectrum
(d) band emission spectrum
Show Answer & Explanation

Answer: (b) continuous emission spectrum

Explanation:
1. A hot incandescent flame with solid particles emits a continuous spectrum.

Teacher's Note:
1. Hot solids and liquids give continuous spectra.

Question 10: Two energy levels of an electron in an atom are separated by 2.3 eV. The frequency of radiation emitted when the electrons go from higher to lower level is
(a) \( 6.95\times10^{14} \) Hz
(b) \( 3.68\times10^{15} \) Hz
(c) \( 5.5\times10^{14} \) Hz
(d) \( 9.11\times10^{15} \) Hz
Show Answer & Explanation

Answer: (c) \( 5.5\times10^{14} \) Hz

Explanation:
1. \( \nu=\dfrac{E}{h}=\dfrac{2.3\times1.6\times10^{-19}}{6.63\times10^{-34}}=5.5\times10^{14} \) Hz.

Teacher's Note:
1. Convert eV to joule.

Question 11: Of the following transitions in the hydrogen atom, the one which gives an emission line of the highest frequency is
(a) \( n=1 \) to \( n=2 \)
(b) \( n=2 \) to \( n=1 \)
(c) \( n=3 \) to \( n=10 \)
(d) \( n=10 \) to \( n=3 \)
Show Answer & Explanation

Answer: (b) \( n=2 \) to \( n=1 \)

Explanation:
1. Emission means falling to a lower level, so (a) and (c) are out. \( 2\to1 \) releases 10.2 eV and \( 10\to3 \) only about 1.4 eV.

Teacher's Note:
1. Highest frequency means largest energy released.

Question 12: For an electron in the second orbit of Bohr's hydrogen atom, the moment of linear momentum is
(a) \( n\pi \)
(b) \( 2\pi h \)
(c) \( \dfrac{2h}{\pi} \)
(d) \( \dfrac h\pi \)
Show Answer & Explanation

Answer: (d) \( \dfrac h\pi \)

Explanation:
1. The moment of linear momentum is angular momentum \( \dfrac{nh}{2\pi}=\dfrac{2h}{2\pi}=\dfrac h\pi \).

Teacher's Note:
1. Angular momentum is quantised.

Question 13: Ionisation energy of He\(^+\) ion at minimum position is
(a) 13.6 eV
(b) 27.2 eV
(c) 54.4 eV
(d) 68.0 eV
Show Answer & Explanation

Answer: (c) 54.4 eV

Explanation:
1. \( E=13.6\,Z^2/n^2 \) with \( Z=2 \), \( n=1 \): 54.4 eV.

Teacher's Note:
1. The minimum position is the ground state.

Question 14: Which of the following transition gives the photon of minimum frequency?
(a) \( n=2\) to \( n=1 \)
(b) \( n=3\) to \( n=1 \)
(c) \( n=3\) to \( n=2 \)
(d) \( n=4\) to \( n=3 \)
Show Answer & Explanation

Answer: (d) \( n=4\) to \( n=3 \)

Explanation:
1. Gap sizes: 10.2, 12.1, 1.89 and 0.66 eV respectively, so \( 4\to3 \) is smallest.

Teacher's Note:
1. Minimum frequency needs the smallest energy gap.

Question 15: The wavelength of radiation emitted is \( \lambda_0 \) when an electron jumps from the third to the second orbit of hydrogen atom. For the electron jump from the fourth to the second orbit of hydrogen atom, the wavelength of radiation emitted will be
(a) \( \dfrac{16}{25}\lambda_0 \)
(b) \( \dfrac{20}{27}\lambda_0 \)
(c) \( \dfrac{27}{20}\lambda_0 \)
(d) \( \dfrac{25}{16}\lambda_0 \)
Show Answer & Explanation

Answer: (b) \( \dfrac{20}{27}\lambda_0 \)

Explanation:
1. \( \dfrac1{\lambda_0}=R\left(\tfrac14-\tfrac19\right)=\dfrac{5R}{36} \), \( \dfrac1{\lambda}=R\left(\tfrac14-\tfrac1{16}\right)=\dfrac{3R}{16} \). \( \dfrac{\lambda}{\lambda_0}=\dfrac{5/36}{3/16}=\dfrac{20}{27} \).

Teacher's Note:
1. Use the Rydberg formula for both jumps.

Question 16: For hydrogen atom electron in nth Bohr orbit, the ratio of radius of orbit to its de-Broglie wavelength is
(a) \( \dfrac n{2\pi} \)
(b) \( \dfrac{n^2}{2\pi} \)
(c) \( \dfrac1{2\pi n} \)
(d) \( \dfrac1{2\pi n^2} \)
Show Answer & Explanation

Answer: (a) \( \dfrac n{2\pi} \)

Explanation:
1. Bohr's condition gives \( 2\pi r=n\lambda \), so \( \dfrac r\lambda=\dfrac n{2\pi} \).

Teacher's Note:
1. The orbit holds n de Broglie wavelengths.

Question 17: According to Bohr's theory of hydrogen atom, for the electron in the nth allowed orbit: I. linear momentum is proportional to \( 1/n \); II. radius is proportional to n; III. kinetic energy is proportional to \( 1/n^2 \); IV. angular momentum is proportional to n. Choose the correct option.
(a) I, III, IV are correct
(b) only I is correct
(c) I and II are correct
(d) only III is correct
Show Answer & Explanation

Answer: (a) I, III, IV are correct

Explanation:
1. \( v\propto1/n \) (I true), \( r\propto n^2 \) (II false), \( KE\propto1/n^2 \) (III true), \( L=nh/2\pi \) (IV true).

Teacher's Note:
1. Radius goes as \( n^2 \), not n.

Question 18: The shortest wavelength which can be obtained in hydrogen spectrum is (\( R=10^7 \) m\(^{-1}\))
(a) 1000 \( \text{\AA} \)
(b) 800 \( \text{\AA} \)
(c) 1300 \( \text{\AA} \)
(d) 2100 \( \text{\AA} \)
Show Answer & Explanation

Answer: (a) 1000 \( \text{\AA} \)

Explanation:
1. The shortest wavelength is the Lyman limit: \( \lambda=\dfrac1R=10^{-7} \) m \( =1000 \) \( \text{\AA} \).

Teacher's Note:
1. Series limit of Lyman is \( 1/R \).

Question 19: In a hydrogen atom, the electron is making \( 6.6\times10^{15} \) rev s\(^{-1}\) around the nucleus in an orbit of radius 0.528 \( \text{\AA} \). The magnetic moment (A m\(^2\)) will be
(a) \( 1\times10^{-15} \)
(b) \( 1\times10^{-10} \)
(c) \( 1\times10^{-23} \)
(d) \( 1\times10^{-27} \)
Show Answer & Explanation

Answer: (c) \( 1\times10^{-23} \)

Explanation:
1. \( I=ef=1.6\times10^{-19}\times6.6\times10^{15}=1.06\times10^{-3} \) A. \( M=I\pi r^2=1.06\times10^{-3}\times\pi\times(0.528\times10^{-10})^2\approx9\times10^{-24}\approx10^{-23} \) A m\(^2\).

Teacher's Note:
1. Magnetic moment is current times loop area.

Question 20: The radius of hydrogen atom in its ground state is \( 5.3\times10^{-11} \) m. After collision with an electron it is found to have a radius of \( 21.2\times10^{-11} \) m. What is the principal quantum number n of the final state of atom?
(a) \( n=4 \)
(b) \( n=2 \)
(c) \( n=16 \)
(d) \( n=3 \)
Show Answer & Explanation

Answer: (b) \( n=2 \)

Explanation:
1. \( r\propto n^2 \): \( n^2=\dfrac{21.2}{5.3}=4 \Rightarrow n=2 \).

Teacher's Note:
1. Radius scales as \( n^2 \).

Question 21: If an electron is revolving around the hydrogen nucleus at a distance of 0.1 nm, what should be its speed?
(a) \( 2.188\times10^6 \) m s\(^{-1}\)
(b) \( 1.094\times10^6 \) m s\(^{-1}\)
(c) \( 4.376\times10^6 \) m s\(^{-1}\)
(d) \( 1.59\times10^6 \) m s\(^{-1}\)
Show Answer & Explanation

Answer: (d) \( 1.59\times10^6 \) m s\(^{-1}\)

Explanation:
1. The ground orbit has \( r_1=0.529 \) \( \text{\AA} \) and \( v_1=2.188\times10^6 \) m s\(^{-1}\). For Coulomb motion \( v\propto r^{-1/2} \): \( v=2.188\times10^6\sqrt{0.529/1}=1.59\times10^6 \) m s\(^{-1}\).

Teacher's Note:
1. For circular Coulomb orbits \( v^2\propto1/r \).

Question 22: Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength \( \lambda \). If R is the Rydberg constant, the principal quantum number n of the excited state is
(a) \( \sqrt{\dfrac{\lambda R}{\lambda R-1}} \)
(b) \( \sqrt{\dfrac{\lambda}{\lambda R-1}} \)
(c) \( \sqrt{\dfrac{\lambda R^2}{\lambda R-1}} \)
(d) \( \sqrt{\dfrac{\lambda R}{\lambda-1}} \)
Show Answer & Explanation

Answer: (a) \( \sqrt{\dfrac{\lambda R}{\lambda R-1}} \)

Explanation:
1. \( \dfrac1\lambda=R\left(1-\dfrac1{n^2}\right)\Rightarrow\dfrac1{n^2}=\dfrac{\lambda R-1}{\lambda R}\Rightarrow n=\sqrt{\dfrac{\lambda R}{\lambda R-1}} \).

Teacher's Note:
1. Solve the Rydberg formula for n.

Question 23: Continuous emission spectrum is produced by
(a) incandescent electric lamp
(b) mercury vapour lamp
(c) sodium vapour lamp
(d) polyatomic substances
Show Answer & Explanation

Answer: (a) incandescent electric lamp

Explanation:
1. A glowing solid filament emits all wavelengths, a continuous spectrum; vapour lamps give line spectra.

Teacher's Note:
1. Hot solids give continuous spectra.

Question 24: The ratio of areas of the electron orbits for the first excited state and the ground state for the hydrogen atom is
(a) 4:1
(b) 16:1
(c) 8:1
(d) 2:1
Show Answer & Explanation

Answer: (b) 16:1

Explanation:
1. Area \( \propto r^2\propto n^4 \): \( \dfrac{2^4}{1^4}=16 \).

Teacher's Note:
1. The first excited state has \( n=2 \).

Question 25: In the Bohr's model of the hydrogen atom, the lowest orbit corresponds to
(a) infinite energy
(b) maximum energy
(c) minimum energy
(d) zero energy
Show Answer & Explanation

Answer: (c) minimum energy

Explanation:
1. The \( n=1 \) orbit has the most negative energy (\( -13.6 \) eV), the minimum energy.

Teacher's Note:
1. Energy rises towards zero as n increases.

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