12th Grade > Physics
ATOMS AND NUCLEI MCQs
Atoms, Nuclei (12th Grade)
Total Questions : 60
| Page 2 of 6 pages
Answer: Option A. -> 3.3×1015HZ
:
A
c=νλ ; ν=cλ=3×108911×10−10
ν=3.3×1015HZ
:
A
c=νλ ; ν=cλ=3×108911×10−10
ν=3.3×1015HZ
Answer: Option B. -> 4862˙A
:
B
1λ1∝[122−132]
1λ1∝536 . . . .(1)
1λ2∝[14−116]
1λ2∝[316] . . . .(2)
λ2λ1=5×1636×3=2027
λ2=2027×6564=4862˙A
:
B
1λ1∝[122−132]
1λ1∝536 . . . .(1)
1λ2∝[14−116]
1λ2∝[316] . . . .(2)
λ2λ1=5×1636×3=2027
λ2=2027×6564=4862˙A
Answer: Option D. -> a0150
:
D
r∝1m ; r2r1=m1m2
r1=a0 ; m1=me ; m2=150me
r2=me150mea0=a0150
:
D
r∝1m ; r2r1=m1m2
r1=a0 ; m1=me ; m2=150me
r2=me150mea0=a0150
Answer: Option A. -> E
:
A
Ionisation energy of atom in nth state EnαZ2n2
For hydrogen atom in ground state (n = 1) and Z = 1 ⇒E=E0 ....... (i)
For Li++ atom in 2nd excited state n = 3 and Z = 3, hence E′=E032×32=E0 ........ (ii)
From equation (i) and (ii) E′=E
:
A
Ionisation energy of atom in nth state EnαZ2n2
For hydrogen atom in ground state (n = 1) and Z = 1 ⇒E=E0 ....... (i)
For Li++ atom in 2nd excited state n = 3 and Z = 3, hence E′=E032×32=E0 ........ (ii)
From equation (i) and (ii) E′=E
Answer: Option C. -> n = 4
:
C
Let electron absorbing the photon energy reaches to the excited state n. Then using energy conservation
⇒−13.6n2=−13.6+12.4⇒−13.6n2=−1.2⇒n2=13.61.2=12⇒n=3.46≃4
:
C
Let electron absorbing the photon energy reaches to the excited state n. Then using energy conservation
⇒−13.6n2=−13.6+12.4⇒−13.6n2=−1.2⇒n2=13.61.2=12⇒n=3.46≃4
Answer: Option C. -> 23.6 MeV
:
C
1H2+1H2→2He4+Q
Total binding energy of helium nucleus = 4×7=28MeV
Total binding energy of each deutron = 2×1.1=2.2MeV
Hence energy released = 28−2×2.2=23.6MeV
:
C
1H2+1H2→2He4+Q
Total binding energy of helium nucleus = 4×7=28MeV
Total binding energy of each deutron = 2×1.1=2.2MeV
Hence energy released = 28−2×2.2=23.6MeV
Answer: Option A. -> 10−14m
:
A
From the options, only at a distance of 10−14m the nuclear force will act. This is equal to ten Fermis, as 1 Fermi =10−15m, and this falls within the short range within which it acts.
:
A
From the options, only at a distance of 10−14m the nuclear force will act. This is equal to ten Fermis, as 1 Fermi =10−15m, and this falls within the short range within which it acts.
Answer: Option A. -> m
:
A
The mass of nucleus formed is always less than the sum of the masses of the constituent protons and neutrons i.e. m<(A−Z)mn+Zmp.
:
A
The mass of nucleus formed is always less than the sum of the masses of the constituent protons and neutrons i.e. m<(A−Z)mn+Zmp.
Answer: Option C. -> 40×1021
:
C
By using P=Wt=n×Et where n = Number of uranium atom fission and E = Energy released due to each fission so 300×106=n×170×106×1.6×10−193600⇒n=40×1021
:
C
By using P=Wt=n×Et where n = Number of uranium atom fission and E = Energy released due to each fission so 300×106=n×170×106×1.6×10−193600⇒n=40×1021
Answer: Option B. -> 87.5%
:
B
N=N0(12)tT1/2. Hence fraction of atoms decayed
=1−NN0=1−(12)tT1/2=1−(12)3×6060=78
In percentage it is 78×100=87.5%
:
B
N=N0(12)tT1/2. Hence fraction of atoms decayed
=1−NN0=1−(12)tT1/2=1−(12)3×6060=78
In percentage it is 78×100=87.5%
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