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Physics NEET MCQ
Atom And Nucleus Mcq
Quiz 6
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Q.1
) Radio carbon dating is done by estimating in the specimen
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a)the amount of oridinary carbon still present
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b) the radio of the amounts of 14 C 6 + 12 C 6
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c) the amount of radio carbon still Present
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d) none of these
Explanation
Answer: (c)
Q.2
) The wave length of second line of Balmer series is 486.4 nm. what is the wave length of the first line of lyman saries ?
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a) 364.8 nm
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b) 729.6 nm
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c) 121.6 nm
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d) None of these
Explanation
For wave length of 2nd line Balmer series Answer: (c)
Q.3
) The innermost orbit of the hydrogen atom has a radius 0.53 Å. what is radius of 2nd orbit is ?
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a) 2.12 Å
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b) 1.06 Å
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c) 21.2 Å
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d) 10.6 Å
Explanation
Answer: (a)
Q.4
) If a hydrogen atom emits a Photon of wave length, the recoil speed of the atom of mass m is given by
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a) h / mλ
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b) mh / λ
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c)mhλ
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d)mλ/h
Explanation
( According comservation of momentum,momentum of photon=momentum of rocil atom. h / λ=m vv=h / mλ Answer: (a)
Q.5
) A freshly Prepared radio active source of half time 2 hr emits raditation of intenisity which is 64 times the Permissible safe level. The minimum time after which is would be possible to work safely with this source is.
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a)6 h
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b) 24 h
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c)12 h
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d)36 h
Explanation
Answer:(c)
Q.6
) In Rutherford experiment, the numer of Particles sccttered at 900 angle are 28 Per min. then the number of Particles at the angle 1200 in Per min will be
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a) 25
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b) 12.0
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c) 50
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d) 112
Explanation
Number of scattering of α -particle at angle θ is given by Answer: (b)
Q.7
) The Rutherford revolution per second made by an electron in the first Bohr orbit of hydrogen atom is of the order of
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a)1015
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b) 1020
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c)1010
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d)1019
Explanation
Answer: (a)
Q.8
) The half time of a radioactive substance is 20 min, difference between points of time when it is 33% disintegeated and 67% dissintenated is approximately
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a) 10 min
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b) 20 min
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c)40 min
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d)30 min
Explanation
Answer: (b)
Q.9
) The size of the atom is of the order of
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a) 10-14 m
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b) 10-10 m
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c) 10-8 m
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d) 10-6 m
Explanation
Answer:(b)
Q.10
) The size of the nucleus is of the order of
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a) 10-10 m
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b) 10-14 m
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c) 10-19 m
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d) 10-3 m
Explanation
Answer: (b)
Q.11
) The ratio of atomic volume of nucleas to volume is of the order of
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a)10-15
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b) 10-10
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c)1015
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d)1010
Explanation
Answer: (c)
Q.12
) Nucleon is common name for
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a) electron and neutron
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b) Proton and neutron
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c) neutron and Positron
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d)neufrom and neurtino
Explanation
Answer: (b)
Q.13
) The nucler 7 N 14 and 6 C 13 can be desribed as
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a)Isotones
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b)Isobars
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c)Isotope
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d)Isomer
Explanation
Answer:(a)
Q.14
) Plutonium decays with half life time 24000 yrs. if Plutonium is stored after 72000 yrs,the fraction of it that remain
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a) 1/2
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b) 1/9
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c) 1/12
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d) 1/8
Explanation
Answer: (d)
Q.15
) The ratio of minimum to maximum wave length in Balmer series is
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a)1 / 4
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b) 5 / 36
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c)3 / 4
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d)5 / 9
Explanation
Answer: (d)
Q.16
) If the binding energy of electron in a hydrogen atom is 13.6 eV , the energy required to remove the electron from the second excited state of Li ++ is
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a) 13.6 eV
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b) 3.4 eV
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c)30.6 eV
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d)122.4 eV
Explanation
For any atom state En=-13.6 Z 2 / n2 for 2nd excitted state n=3 , for Li Z=3 ∴ En=(-13.6) × (3)2 / 32 ∴E2=13.6Answer: (a)
Q.17
) A radioactive substance decays to(1 /16) th of its initial activity in 40 days. the half life of the radioactive substance expressed in day is
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a)20
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b) 5
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c)10
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d)4
Explanation
Answer:(c)
Q.18
) A complete the reaction
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a) 36 Kr 90
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b) 36 Kr 89
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c) 36 Kr 91
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d) 36 Kr 92
Explanation
Answer: (d)
Q.19
) 6 C12 absorbs an energetic neutron and emits a β Partical. The resulting nucleus is
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a)7 N13
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b)7 N14
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c)6 N 13
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d)6 C12
Explanation
Answer: (a)
Q.20
) Two deutrons each of mass m fuse to form helium resulting in release of energy E the mass of helium formed is
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a) m + E /c2
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b) E / mc2
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c)2m - E / c2
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d)2m + E / c2
Explanation
Answer: (c)
Q.21
) what Percent of original radio active substans is left after 5 half life time
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a) 3 %
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b) 5%
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c)6%
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d)12%
Explanation
Answer:(a)
Q.22
) In which region of eletromagnetic spectum does the Lyman series of hydrogen atom like
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a) x-ray
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b) Intrared
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c) visible
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d) ultraviolet
Explanation
Answer: (d)
Q.23
) In terms of Rydergi constent R. The wave number of first Balmer line is
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a)5R / 36
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b) 8R / 9
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c) R
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d)8R / 20
Explanation
Answer: (a)
Q.24
) The hydrogen atom can give spectral lines in the series Lyman, Balmer and Paschen. which of the following statement is correct
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a) Lyman series is in the intrared region
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b) Balmer series is in the ultravioet region
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c)Balmer series is in the visible region
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d)Paschen series is in the visible region
Explanation
Answer: (c)
Q.25
) The ionization energy of hydrogen atom is 13.6 eV. The ionization energy of helium atom would be
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a) 27.2 eV
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b) 13.6 eV
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c)54.4 eV
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d)6.8 eV
Explanation
Ionigation energy of atom E=13.6Z2 /n2 evFor helum Z=2 E=13.6 × 4=54.4 eV Answer:(c)
Q.26
) A gamma ray photon creates an electon- Positon Pair. If the rest mass energy of an electron is 0.5 MeV. and the total kinetrc energy 0.7 MeV, then the energy of the gamma ray Photon must be
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a) 3.9 MeV
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b) 1.78 MeV
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c) 0.78 MeV
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d) 0.16 MeV
Explanation
The energy of X - ray=Rest mass energy + K. Eenergy of X - ray=(0.5 + 0.5 ) + 0.78energy of X - ray=1.0 + 0.78energy of X - ray=1.78 eVAnswer: (b)
Q.27
) The masses of netron and Proton are 1.0087 amu and 1.0073 amu respectively. It the neutron and Protons combins to form binding energy of the helium nucleus will be (Mass of Helium = 4.0015amu
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a)14.2 MeV
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b) 28.4 MeV
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c)27.3 MeV
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d)20.8 MeV
Explanation
Total protron in helium is 2 and neutron in helium is 2 Total mass before combination = 2×1.0087 +2 ×1.0073 = 4.032 amu Mass of Helium = 4.0015 amu ∴ mass defect ΔM=2(4.032)−4.0015= - 0.0305amu 0.0305amu mass converted to energy to hold the nucleons Now 1 amu = 931 Mev of energy Binding energy E=931ΔM=28.4MeV Answer: (b)
Q.28
) Large angle scattering of α - particle could not be explained by
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a) Thomson modal
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b) Ruther Rutherford and Thpmoson modal
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c)Both rutherford and Thomoson modal
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d) Neither Ruthorford modal nor Thomson modal
Explanation
Answer: (a)
Q.29
) The energy of an elecorn in nth orbit of hydrogen is 13.6 /n 2 eV. Energy required to exite the electron form the first orbit 4th orbit is
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a)13.6
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b) 3.4 eV
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c)0.85 eV
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d)1.5 eV
Explanation
Answer:(c)
Q.30
) The activity of a radioactive sample is measured as no conuts per minute at t=o and No counts per minute t=5 min. The time (in min) at which activity reduces to half its value is
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a) log e(2/5)
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b) 5 log 10 2
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c) 5 log e 2
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d) log 10 (2/5)
Explanation
Answer: (c)
0 h : 0 m : 1 s
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