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Atom And Nucleus Mcq
Quiz 1
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Q.1
) Electron cannot be constituent of the nucleus because [ AFMC 2010]
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a) the electron is a negatively charged particle
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b) the compton wave length of electron is larger than the size of nucleus
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c) the mass of the electron is very small as compared to the mass of the neutron or proton
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d) None of these
Explanation
Electrons are negatively charged, so if electrons be in nucleus then electron proton can neutraize each other. Answer: (a)
Q.2
7) The energy of the ground electronic state of hydrogen atom is - 13.6 eV. The energy of the first excited state will be{AFMC 2001}
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a)-- 52.4 eV
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b)- 27.2 eV
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c)- 68 eV
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d)- 3.4 eV
Explanation
Answer:(d)
Q.3
) What is the de- Broglie wave length of 1 kg mass moving with velocity of 10 m/ s ?{AFMC 2001}
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a) 6.626 × 10 -35 m
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b) 6.626 × 10 -33 m
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c) 6.626 × 10 -34 m
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d) none of these
Explanation
Answer: (a)
Q.4
) Which of the following phenomenon shows the transverse nature of light ?{AFMC 2001}
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a) Photo eletric effect
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b) Interference
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c)Polarization
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d)Difraction
Explanation
Polarixation can happen only in case of transverse wave. So, this phenominon shows transverse nature of light.Answer: (c)
Q.5
0) The larger scale destruction, that would be caused due to use of nuclear weapons is known as{AFMC 2004}
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a) neutron holocaust
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b) thermonuclear reaction
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c)neutron reproduction factor
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d)none of these
Explanation
It is known as Neuclear holocaust.Answer: (a)
Q.6
1) According to classical theory of Rutherford model, the path of electron will be{AFMC 2003}
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a) parabolic
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b) hyperbolic
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c)circular
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d)elliptical
Explanation
According to Rutherford model, an atom have a positevely charged nucluse and eletrons revolves around the nucleus in circular path. Answer:(c)
Q.7
) The total energy of the elcton in the first excited state of hydrogen is -3.4 eV. what is the kinatic energy of the electron in this state?
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a) 6.8 eV
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b) 3.4 eV
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c)-3.4 eV
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d) -6.8 eV
Explanation
Answer:(b)
Q.8
The velocity of an electron in the inner most orbit of an atom is ...[ AFMC 1997]
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a)highest
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b) lowest
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c)mean
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d)zero
Explanation
As the radius decreases orbital speed increases then option (a) is correctAnswer: (a)
Q.9
The mass number of He is 4 and that of sulphur isThe radius of sulphur nucleus is larger than that of helium by a factor of ...[ AFMC 1997]
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a) 2
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b) 4
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c)√8
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d)8
Explanation
Radius of nucleus r=R0 A1/3Atomic mass number A for helium=4 Atomic mass number for sulphur=32By substituting values and taking the ratio of Radius of Sulphur to Radius of Helium we gateRatio=(32/4) 1/3=2 Answer: (a)
Q.10
Energy produced in sun is due to ... [ AFMC 1998]
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a) motion of electrons and ion
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b) chemical reaction
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c)fusion reaction
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d)fission reaction
Explanation
Answer:(c)
Q.11
The half life of a radio active substance is 3.6 days. How much of 20 mg of that radioactive substance will remain after 36 days
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a) 0.0019 mg
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b) 1.109 mg
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c) 1.019 mg
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d) 0.019 mg
Explanation
36 days=36 /3.6=10 half life Substance left after 10 half life=20 / (2 10=0.019 mg or divide 20 by 2 10 times Answer: (d)
Q.12
The large scale destruction, that would be caused due to the use of nuclear weapons is known as.. [ AFMC 1999]
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a)neutron reproduction factor
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b) nuclear holocaust
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c)thermo nuclear reaction
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d)none of these
Explanation
Answer: (b)
Q.13
The thermions are
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a) positron
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b) photons
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c)electrons
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d)protons
Explanation
Thermions are electrons when we heat the object, it starts emitting electrons. These electrons are also known as thermionsAnswer: (c)
Q.14
The energy released by fission of one atom of 92U235 is 200 MeV. The number of fission required per second to produce a power of 1kW is ..[ AFMC 2001]
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a)3.125×109
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b)3.125×1012
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c)3.125×1013
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d)3.125×1011
Explanation
Energy released in joule per fission is=200× 106 ×1.6× 10-19Energy released in joule per fission is=320×10-13 J=320×10-16 kJIf 'n' is the number of fission per second then n=total energy per second / energy per fission n=1kW/ 320×10-16 n=3.125×1013 Answer:(c)
Q.15
A substance reduces to one sixteenth of its original mass in 2 hours. The half life period of the substance will be..[ AFMC 2000]
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a) 30 min
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b) 90 min
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c) 45 min
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d) 60 min
Explanation
by formula so, 2 hours involves 4 half lifeHalf life=(2×60)/4=30 minutesAnswer: (a)
Q.16
A radio active material decays by simultaneous emission of two particles with respective half life-lives of 1620 year and 810 year. The time, after which one fourth of the materiel remains, is...[ AFMC 2002]
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a)3860 year
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b) 4240 year
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c)2380 year
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d)1080 year
Explanation
The composite half life be τ then Time required for remaining one fourth of material=two half lives=2(540)=1080 yearsAnswer: (d)
Q.17
In fission of U235, the percentage of mass converted into energy is about..[ AFMC 2003]
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a) 0.1%
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b) 0.25%
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c)0.3%
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d)2%
Explanation
Energy released per fission of U235 is 200MeV. During fission one fifth of mass of proton is converted into energy. Thus percentage mass converted to energy is Answer: (a)
Q.18
A radio active element 90X238 decay into 83X222, then the number of β particle emitted are... [ AFMC 2001]
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a) 1
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b) 2
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c)4
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d)6
Explanation
Atomic mass number decrease by=238-222=16number of α particle emitted=16/4=4 Atomic number should have decreased by=4×2=8Actual decrease in atomic number is=90-83=7 Thus it must have emitted one β particle, by converting one neutron to proton so 4 alpha and one β particle Answer:(a)
Q.19
Assuming that about 200 MeV energy is released per fission of 92U235 nuclei. What would be the mass of U-235 consumed per day in the fission reactor of power 1MW approximately
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a) 10 kg
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b) 100kg
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c) 1 g
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d) 10-2g
Explanation
Amount of energy released per fission is 200 MeV=200(1.6×10-13=320-13 Require energy per sec=106 J Number of atoms required per second=required energy / energy released per fission Number of atoms required per second=106 / 320-13 Number of atoms required per second=1 / 32×10-18 Weight of fissile material required per second will be Answer: (c)
Q.20
The spectrum obtained from a sodium vapour lamp is an example of... [ CBSE-PMT 1995]
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a)band spectrum
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b) continuous spectrum
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c)emission pectrum
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d)absorption spectrum
Explanation
A spectrum is observed, when light coming directly from a source is examined witha spectroscope. Therefore spectrum obtained from a sodium vapour lamp is emission spectrumAnswer: (c)
Q.21
Out of the following which is not a possible energy for photon to be emitted by hydrogen atom according to Bohr's atomic model? [ CBSE-PMT 2011]
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a) 1.9 eV
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b) 11.1eV
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c)13.6eV
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d)0.65eV
Explanation
Energy for orbit is given by E=-13.6/n2 n=1 E1=-13.6 eV n=2 E2=-3.4 eV n=3 E3=-1.51eV n=4 E4=-0.85 eV n=5 E5=-0.58eV Thus differenc of 11.1 eV is not possible Answer: (b)
Q.22
The Bohr model of atoms ... [ CBSE-PMT 2004]
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a) Predicts the same emission spectra for all types of atoms
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b) assumes that the angular momentum of electrons is quantised
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c)uses Einstein's photoelectric equation
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d)predicts continuous emission spetra for atoms
Explanation
In Bohrs model anglar momentum is quantisec anular momentum=nh/2π Answer:(b)
Q.23
The total energy of an electron in the first excited state of hydrogen atom is about -3.4eV. Its kinetic energy in this state is .. [ CBSE-PMT 2005]
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a) 3.4 eV
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b) 6.8eV
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c) -3.4eV
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d) -6.8eV
Explanation
Kinetic energy=- Mechanical energy Mechnical energy=-13.6 /n2=-13.6 /22 For first excited state n=2 Mechanical energy=-13.6/4=-3.4eVThus kinetic energy=+3.4 eV Answer: (a)
Q.24
The half life of radium is about 1600 years. If 100 g of radiumm exists now, 25g will remain unchanged after... [ CBSE-PMT 2004]
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a)3200 years
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b) 4800 years
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c)6400 years
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d)2400 years
Explanation
100g will become 25g in two half lives, so it is 3200 yearsAnswer: (a)
Q.25
When hydrogen atom is in its first excited level its radius is .. [ CBSE-PMT 1997]
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a) four times its ground state radius
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b) twice
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c)same
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d)half
Explanation
radius of hydrogen atom R∝ n2 now for first excited state n=2. thus radius of first excited state is four times the radius in ground state.Answer: (a)
Q.26
In terms of Bohr's radius a0, the radius of the second Bohr orbit of Hydrogen atom is given by... [ CBSE-PMT 1992]
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a) 4a0
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b) 8a0
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c)√2 ×a0
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d)2a0
Explanation
radius of hydrogen atom R∝ n2∴ radius of 2nd Bohr's orbit=4a0 Answer:(a)
Q.27
When electron jump from the fourth orbit to the second orbit, one gets the ... [ CBSE-PMT 2000]
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b) second line of Paaschen series
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a) second oline of Lyman series
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c) second line of Balmar series
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d) first line of Pfund series
Explanation
When the electron drop from any orbit to second orbit, then wavelength of line obtained belong to Balmer series. Answer: (c)
Q.28
) The enery difference between the first two levels of hydrogen atom is 10.2 eV. what is the corresponding energy difference for a singly ionized helium atom?
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a) 10.2 eV
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b) 81.6 eV
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c)20.4 eV
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d) 40.8 eV
Explanation
Answer: (d)
Q.29
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 of Z of hydrogen like ion is .. [ CBSE-PMT 2011]
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a)3
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b) 4
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c)1
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d)2
Explanation
According to equation for spectral line emitted by hydrogen like ion is given byFor Lyman series of hydrogen Z=1 thus For second line of Balmer series of Hydrogen like ionGiven λ1=λ2 On simplification Z=2Answer: (d)
Q.30
Which of the following transitions in hydrogen atoms emits the photon of highest frequency? [ CBSE-PMT 2000]
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a) n=2 to n=1
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b) n=2 to n=6
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c)n=6 to n=2
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d)n=1 to n=2
Explanation
Frequency of wave emitted Transmtion from higher m to lower energy level should have least difference to have maximum frequencyin given gase m=2 and n=1 will be the answerAnswer: (a)
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