12th Grade > Physics
ATOMS AND NUCLEI MCQs
Atoms, Nuclei (12th Grade)
Total Questions : 60
| Page 4 of 6 pages
Answer: Option D. -> 1.0×106 disintegrations/sec
:
D
1.0×106 disintegrations/sec
:
D
1.0×106 disintegrations/sec
Answer: Option D. -> 9×1013J
:
D
By using E=Δm.c2⇒E=(0.1100×1)(3×108)2=9×1013J
:
D
By using E=Δm.c2⇒E=(0.1100×1)(3×108)2=9×1013J
Answer: Option D. -> In nuclear fission, energy is released by fragmentation of a very heavy nucleus
:
D
In nuclear fission, energy is released by fragmentation of a very heavy nucleus
:
D
In nuclear fission, energy is released by fragmentation of a very heavy nucleus
Answer: Option C. -> 5→4
:
C
Infra- red radiation has lesser energy than Ulta-Violet radiation. Hence the transition should take place between energy levels which have lesser energy.
5→4
:
C
Infra- red radiation has lesser energy than Ulta-Violet radiation. Hence the transition should take place between energy levels which have lesser energy.
5→4
Answer: Option B. -> 13.6 eV
:
B
KE=-E
K.E. = 13.6 eV
:
B
KE=-E
K.E. = 13.6 eV
Answer: Option B. -> 108.8 eV
:
B
E=−13.6z2n2
E1=−13.6×321=−13.6×9
E3=−13.6×3232=−13.6
E3−E1=−13.6+(13.6×9)
=108.8eV
:
B
E=−13.6z2n2
E1=−13.6×321=−13.6×9
E3=−13.6×3232=−13.6
E3−E1=−13.6+(13.6×9)
=108.8eV
Answer: Option C. -> 3.4 eV
:
C
E2=−13.6n2=−13.622
=−3.4eV
E=3.4eV
:
C
E2=−13.6n2=−13.622
=−3.4eV
E=3.4eV
Answer: Option D. -> – 3.4 eV
:
D
E=−13.6n2n=2
=−13.6n2=−3.4eV
:
D
E=−13.6n2n=2
=−13.6n2=−3.4eV
Answer: Option D. -> 1 : 8
:
D
T1T2=(n1n2)3=(12)3=18
T1:T2=1:8
:
D
T1T2=(n1n2)3=(12)3=18
T1:T2=1:8
Answer: Option C. -> 3.2 m/s
:
C
P=EC=(13.6−3.4)3×1081.6×10−19
=10.2×1.6×10−193×108
V=pm=3.4×1.6×10−191.67×10−27×108=V=3.2ms−1
:
C
P=EC=(13.6−3.4)3×1081.6×10−19
=10.2×1.6×10−193×108
V=pm=3.4×1.6×10−191.67×10−27×108=V=3.2ms−1