PART A – PHYSICS
1.
2
b)
Kinetic energy + Potential energy = Total energy
De-Broglie wavelength
2.
d) the intensities of light incident on a, b and c are all different
Intensity
3.
c) 1
From conservation of linear momentum both the particles will have
equal
and opposite momentum.
The de-Broglie wavelength is given by
4.
b) E
3.4 eV, O | 6.6u 10 10 m
The potential energy
total energy
or
.
Let p = momentum and
mass of the electron.
De Broglie wavelength,
5.
2.176 u 10 18 J
a)
E n1 on2
E
or ,
hc
O
A A
n22 n22
On1on2
so,
On1o n2 1
0.054
n12
1
1
2
n1
n2 1
1 1
n12 n22
1
1.054
n22
n2 1
2
1028
1.054
975
2
Page No - 1 -
n2
so, taking n2
3 gives n1 1
hc § 1
O ¨© 1
so, A
6.
2,3, 4........ so an find integral value of n2
2,3, 4........n1 we find,
1
1·
18
¸ J 2.176 u 10 J
9¹
h
mA
a = A cost t, Z = 1
v = A sin t at t = S/2; v = A
? O = h/mA
a)
7.
9a 2I
b)
4c
Radiation pressure =
Area of hexagon =
8.
a)
2mc O 2
h
P
9.
a)
h
O
kE
P2
2m
h2
2mO 2
hc
O0
1
e
N (t )
=
No
N o e – Ot
No
Probability = e – O
1
O
e –1
1
e
? (a) is correct.
10.
d) two or three atoms
An excited H-atom cannot emit 1.9ev, 10.2ev and 12.1ev simultaneously
11.
b) Targets of high atomic numbers have smaller energy differences as
compared with targets of low atomic numbers.
| E |v Z2
So targets of higher atomic numbers have higher energy differences.
Page No - 2 -
12. b)
41 kV
For shortest wavelength, maximum K.E of electron is converted to X-ray
hc
...(1)
3 eV
o
b
Let Vo be initial voltage
hc
3 e 3Vo
b # 20
1
O 20 1
2
O
3
...(2)
Ÿ 3O 60 O
b 3 30 pm
Vo # 41kV
13. c)
25
v
9
Q
a Z b
14. b) product of slope of the line and charge on the electron
e V0 hX W Where V0 is the stopping potential
15. d) doubly ionized lithium
16. d)
nh
2S mK
F
dU
dr
Force F
K
r
mv 2
also mvr
r
nh
2S
17. b)
photon of energy 10.2eV and electron of energy 1.4eV
18. b)
Z A Z B ; VA ! VB
OminD
19. a)
1
and
V
C
O
a Z b
1.51 eV
Angular momentum
nh
?n
2S
3
20. c)
both a and b are independent of target material
21. d)
None
I Ÿ eb
(where e o electronic charge)
Page No - 3 -
e
22. d)
O t1 t2
Conceptual
23. a)
30
(or) t = 30 min.
3
s
8
24. d)
t
N
No
t
T
25.
§ 1 ·T
¨ ¸
©2¹
1
4
t
2
2 or T
3
S.
8
b) 3 : 2
Ox
Oy
Since,
N xO x
26.
3
4u 2
2
3
N yO y Ÿ
Nx
Ny
3
2
d) 8 × 109 years
Let the initial no be N o
For X, bx 3
Nx
0.693
,
Tx
N o e O xt , N y
by 3
N o e O yt ,
0.693
Ty
Nx
Ny
20
80
1
4
27. a) 4.95 Mev
Total B.E. of C13- Total B.E. of C12 = (13 x 7.47) – 12 x 7.68
28. b) 180 MeV
‡
B.E products B.Ereac tan ts
Page No - 4 -
29. b) 6.5 MeV
Mass of one atom of U235 is
235.121420 a.m.u.
Mass of one neutron = 1.008665 a.m.u.
Sum of the masses of U235 and neutron
=236.130085 = 236.130 a.m.u.
Mass of one atom of U236 is
236.123050 a.m.u. = 236.123 a.m.u.
Mass defect = 236.136 – 236.123
= 0.007 a.m.u.
Therefore, energy required to remove one neutron is
0.007 × 931 MeV = 6.517 MeV = 6.5 Mev
3 ln 2
2t
Probability that either decrees
30. b)
Page No - 5 -