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Physics NEET MCQ
Atom And Nucleus Mcq
Quiz 2
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Q.1
the energy of hydrogen atom in nth orbit is En then the energy in the nth orbit of a single ionised helium atom will be.. [ CBSE-PMT 2001]
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a) 4En
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b) En /4
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c)2En
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d)En / 2
Explanation
Energy of nth orbit En ∝ Z2 / n2 For helium Z=2∴ Energy of nth orbit of Helium En=4En Answer:(a)
Q.2
The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emitt radiations of 6 wavelengths. Maximum wave length of emitted radiation corresponds to the transition between... [ CBSE-PMT 2009]
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a) n=3 to n=1 satates
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b) n=2 to n=1 states
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c) n=4 to n=3 states
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d) n=3 to n=2 states
Explanation
Number of wave lenths=n(n-1) /2 Here n is the energy state ∴ 6=n(n-1) / 2 n=4 thus answer is n=4 to n=3 Answer: (c)
Q.3
In which of the following systems will the radius of the first orbit (n=1) be minimum? [ CBSE-PMT 2003]
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a)Hydrogen atom
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b) Doubly ionized lithium
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c)Singly ionized helium
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d)Deuterium atom
Explanation
radius R∝ 1/ Z For lithium Z=3 thus Li+2Answer: (b)
Q.4
Ionization potential of hydrogen atom is 13.6eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1eV. According to Bohr's theory, the spectral lines emitted by hydrogen will be.. [ CBSE-PMT 2006]
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a) three
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b) four
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c)one
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d)two
Explanation
Energy of ground state is 13.6 eV and excited by radiation of 12.1eV . Thus excited satate enegy=-13.6+12.1=1.5 eVEn=-13.6 / n2From above formula ,Energy of third energy level is -1.5 Thus posibe transitions are 1→2, 1→3, 2→3so three lines are possibleAnswer: (a)
Q.5
An alpha nucleus of energy ½ m v2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to .. [ CBSE-PMT 2010]
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a) 1 / Ze
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b) v2
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c)1/m
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d)1/v4
Explanation
Kinetic energy of alpha nucleus is equall to electrostatic potential energy of the system of the alpha particle and the heavy nucleus That is Here r is the distance of closestt approach2e is charge on alpha particleZ is atomic number of elementOn solving above equation for 'r' we getsince e, m, Z, k are constant r ∝ 1/v2 Answer:(c)
Q.6
Which source is associated with a line emission spectrum? [ CBSE-PMT 1993]
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a) Electric fire
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b) Neon street sign
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c) Red traffic light
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d) sun
Explanation
A neon tube has low-density neon gas in it that is energized by the electric current so it produces an emission spectrum. The light produced by a red neon sign is due to the emission of light by excited neon atoms. wavelength Answer: (b)
Q.7
An electron in the hydrogen atom jumps from excited state n to the ground state. The wave length so emitted illuminates a photosenseitive material having work function 2.75eV. If the stopping potential of the photoelectron is 10V, the value of n is .. [ CBSE-PMT 2011]
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a)3
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b) 4
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c)5
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d)2
Explanation
KE=10EvΦ=2.75 eVTotal incident energy E=Φ+KE=12.75eVTherefore energy is released when electron jumps from the excited state n to the ground state ∴ E4 - E1=-0.85 - (-13.6) eVE4 - E1=12.75 eVn=4Answer: (b)
Q.8
When a hydrogen atom is raised from ground state to an excited state.. [ CBSE-PMT 1995]
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a) P.E. decrases and KE increases
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b) P.E. increases and K.E. decreases
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c)both K.E. and P.E. decreases
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d)absorption spectrum
Explanation
Here 'r' is the radius of the orbit which increases as we move from ground to an excited state. Therefore, when a hydrogen atom is raised from the ground state, it increases the value of 'r'. As a result of this P.E. increases ( decrease in negative) and K.E. decreasesAnswer: (b)
Q.9
What is the rdius of iodine atom ( At.No 53, mass number 126)...[CBSE-PMT 1988]
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a) 2.5×10-11 m
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b) 2.5×10-9 m
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c)7×10-9 m
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d)7 × 10-6 m
Explanation
53 electrons in iodine artom are distributed as 2,8,18,7 ∴ n=5 and Z=53formula for radius is Answer:(a)
Q.10
The ionisation energy of hydrogen atom is 13.6 eV, the ionisation energy of helium atom would be .. [ CBSE-PMT 1988]
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a) 13.6 eV
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b) 27.2 eV
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c) 6.8 eV
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d) 54.4 eV
Explanation
E ∝Z2 for helium Z=2 ∴ EHe=4× (2)2 EHe=2×13.6=54.4eV Answer: (d)
Q.11
The radius of hydrogen atom in its ground state is 5.3×10-11 m. After collision with an electron it is found o have a radius of 21.2×10-11 m. What is the principal quantum number n of the final state of the atom.. [ CBSE-PMT 1994]
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a) n=4
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b) n=2
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c)n=16
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d)n=3
Explanation
r ∝ n2 ∴ radius of final state / radius of initial state=n2 Answer: (b)
Q.12
Curi is a unit of .. [ CBSE-PMT 1989]
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a) energy of gamma ray
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b) half-life
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c)radioactivity
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d)intensity of gamma-rays
Explanation
Answer: (c)
Q.13
Fusion reaction takes place at high temperature because ... [ CBSE-PMT 2011]
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a) nuclei break up at high temperature
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b) atoms get ionisec at high temperature
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c)kinetic energy is high enough to overcome the coulomb repulsion between nuclei
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d)molecules break up at high temperature
Explanation
When the coulomb repulsion between the nuclei is overcome then nuclear fusion reaction takes place. This is possible when temperature is too high Answer:(c)
Q.14
Which of the following statement is true for nuclear forces?... [ CBSE-PMT 1990]
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a) they obey the inverse square law of distance
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b) they obey the inverse third power law of distance
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c) they are short range forces
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d) they are equal in strength to electromagnetic forces
Explanation
Nuclear forces are short range attractive forces which balance the repulsive forces between the protons inside the nucleus. Answer: (c)
Q.15
Heavy water is used a moderator in a nuclear reactor. The function of the moderator is .. [ CBSE-PMT 1994]
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a)to control energy released in the reactor
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b) to absorb neutrons and stop chain reaction
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c)to cool the reactor
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d)to slow down the neutrons to thermal energies
Explanation
Answer: (d)
Q.16
An electron with ( rest mass mo) moves with a speed of 0.8c. Its mass when it moves with this speed is .. [ CBSE-PMT 1991]
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a) mo
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b) mo /6
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c)5mo /3
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d)3mo /5
Explanation
Answer: (c)
Q.17
Fission of nuclei is possible because the binding energy per nucleon is them .. [ CBSE-PMT 2005]
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a)increases with mass number at low mass number
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b) decreases with mass number at lower mass number
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c)increases with mass number at high mass number
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d)decreases with mass number at high mass number
Explanation
Binding energy per nucleon is smaller for lighter as will as heaviour nucleus. But fusion reaction occure for small mass number nuclei and fission reaction occures for large mass number nuclei to attain reaction binding energy per nucleon. Answer:(d)
Q.18
The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is .. [ CBSE-PMT 2010]
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a) 30.2 MeV
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b) 23.6MeV
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c) 2.2MeV
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d) 28.0MeV
Explanation
Binding energy of two deuterium nuclei=2(1.1×2)=4.4 MeV Binding energy of one helium nucleus=4×7.0=28.0MeV∴ Energy released=28-4.4=23.6MeV Answer: (b)
Q.19
2) The work function of a metallic substance is 5 eV. The threshold frequency is approximately{AFMC 2009}
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a) 1.6 × 10 7 Hz
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b) 8.68 × 10 15 Hz
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c) 9.68 × 10 17 Hz
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d) 1.2 × 10 15 Hz
Explanation
Work function (W 0 ) and the threshold frequency (v0) are related as below Answer: (d)
Q.20
The half life of a radio active isotopes 'X' is 50 years. It decays to another element 'Y' which is stable. the two elements 'X' and 'Y' were found to be in the ratio of 1:15 in a sample of a given rock. the age of the rock was estimated to be.. [ CBSE-PMT 2011]
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a)150 years
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b) 200 years
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c)250 years
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d)100 years
Explanation
Let number of atoms in X=Nx Let number of atoms of Y be=Ny By equation Initially sample was made up of X but decays to Y after passage of time, out of 16 parts one part is Nx Now 16=24 So, total 4 half lives are passed, so, age of rock is 4×50=200 yearsAnswer: (b)
Q.21
Two nuclei have their mass numbers in the ratio of 1:The ratio of their nuclear densities would be .. [ CBSE-PMT 2008]
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a) 1:3
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b) 3:1
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c)31/3 : 1
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d)1:1
Explanation
We know that nulear density for all element is same Answer: (d)
Q.22
A nucleus ruptures into two nuclear parts, which nave their veloity ratio equal to 2:What will be the ratio of their nuclear size ( nuclear radius)?.. [CBSE-PMT 1996]
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a) 21/3 : 1
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b) 1:21/3
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c)31/2 : 1
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d)1: 31/2
Explanation
Applying law of conservation of momentum m1v1=m2v2we know that density of nucleus is constant thus m∝r3 Answer:(b)
Q.23
Two radioactive nucleous P and Q in a given sample decay into a stable nucleus R. At time t=0, number of P species are 4No and that of Q are No. Half life of P is 1 mintes where as that of Q is 2 minutes. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be.. [ CBSE-PMT 2011]
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a) 3No
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b) (9/2)No
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c) (5/2) No
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d) 2No
Explanation
Initilly P=4No , half life is 1 min After first half life P=2No After second half life P=No After third half life P=No/2 after fourt half life P=No/4 Total time=4 minSo R=4No - No/4 R=15No / 4 --(1) Initially Q=No , half life is 2 minAfter first half life Q=No / 2 After second half life Q=No / 4 Total time=4 minSo R=No - No /4R=(3/4)No --(2)form (1) and (2) total R=15No / 4 + (3/4)No Total R=(9/2) No Answer: (b)
Q.24
The activity of radioactive sample is measured as No counts per minutes at t=0 and No/e counts per minute at t=5 minutes. The time ( in munutes) at which the activity reduces to half its value is .. [ CBSE-PMT 2010]
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a)loge(2/5)
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b) 5 / loge2
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c)5 log102
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d)5 loge2
Explanation
N=Noe-λtHere t=5 minutesNo /e=Noe-5λ∴ 5λ=1λ=1/5Now Τ1/2=ln2/λ=5ln2 Answer: (d)
Q.25
A radioactive nucleus of mass M emits a photon of frequency ν and the nucleus recoils. The recoil energy will be.. [ CBSE-PMT 2011]
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a) Mc2 - hν
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b) h2 ν2 / 2Mc2
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c)zero
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d)hν
Explanation
Momentum P=MvMv=E/c=hν / cRecoil energy (K.E) Answer: (b)
Q.26
The mass number of a nucleus is .. [ CBSE-PMT 2003]
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a) sometimes less than and some times more than its atomic number
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b) always less than its atomic number
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c)always more than its atomic number
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d)sometimes equal to its atomic number
Explanation
Incase of Hydrogen atommass Number=Atomic Number Answer:(d)
Q.27
The voume occupied by an atom is greater than the volume of the nucleus by a factor of about.. [ CBSE-PMT 2003]
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a) 1015
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b) 10
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c) 105
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d) 1010
Explanation
We know that V∝ r3 Answer: (a)
Q.28
For a nuclear fusion process, the sitable nuclei are.. [ CBSE-PMT 2002]
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a)any nuclei
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b) heavy nuclei
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c)light nuclei
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d)nuclei lying in the middle of the periodic table
Explanation
Answer: (c)
Q.29
It is possible to understand nuclear fission on te basis of the .. [ CBSE-PMT 2000]
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a) liquid drop model of the nucleus
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b) meson theory of the nuclear forces
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c)proton-proton cycle
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d)independent particle model of the nucleus
Explanation
Answer: (a)
Q.30
After 1 α and 2β emissions.. [ CBSE-PMT 1999]
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a) mass number reduced by 4
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b) mass number reduced by 5
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c)mass number reduced by 6
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d)mass number increased by 4
Explanation
Emission of 1α particle led to decrease in atomic number by 2 while mass number by 4. On the other hand, emission of 2β particles increases atomic number by 2. Henece, overall emission of 1α and 2β particles led to decrease in mass number by 4. Answer:(a)
0 h : 0 m : 1 s
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