Dual Nature of Matter & Radiation Practice Questions - DPP for NEET

 Page 1


1. (a). l
a
 = 
0.101
V
 Å, V = 
0.101
0.004
V = 25.25 V , V = 637.5 V
E
a
 =q
a
 × V
a
 » 1275 eV
2. (d). l = 
h
mv
Q  v = 
8
c 3 10
20 20
´
=
 = 1.5 × 10
7
 m/sec
h = 6.626 × 10
–34
 J-s, m = 1.67 × 10
–27
 kg
\  
34
277
6.626 10
1.67 10 1.5 10
-
-
´
l=
´ ´´
Þ  l = 2.64 × 10
–14
 m
3. (a). Q l µ 
1
m
  Þ l
e
 µ 
e
1
m
, l
p
 µ 
p
1
m
\ 
p
e
pe
m
m
l
=
l
4. (c).  l = 
h
mv
 Þ v = 
h
ml
 ,
34
31 10
6.6 10
v
9.1 10 10 10
-
--
´
=
´ ´´
 = 7.2 × 10
5
 m/s
5. (b). l µ 
1
m
 ,
p
e
pe
m
1836
m1
l
==
l
6. (a). l
p
 = 
pp
h
2meV
Þ l
d
 = 
dd
h
2meV
\  
pp
d
p dd
me
me
l
=
l
   Q  m
d
 = 2m
p
,
e
d
 = e
p
  Þ  
pp
d
p pp
me
1
2me 2
l
==
l
7. (d). K
max
 of photoelectrons doesn't depends upon intensity
of incident light.
8. (b).
  
34
31 19
6.62 10
2 9.1 10 13.6 1.6 10
-
--
´
l=
´´ ´ ´´
Þ  l = 3.3 × 10
–10
 m = 3.3 Å
9. (c).  l = 
12.27
V
 Å
Þ V = 40–20 = 20 V olt
Þ  l = 
12.27
20
 Å = 2.75 Å
10. (d). Wavelength of electrons is l = 
150
V
Å
Now, electrons have energy of 40 KeV , therefore they
are accelerated through a potential difference of 40 ×
10
3
 volt.
       l =  
3
150
40 10 ´
 = 0.061 Å
\  Resolving limit of electron microscope = 0.061 Å
11. (d). The linear momentum of the photon
= 
34
9
h 6.63 10
122 10
-
-
´
=
l
´
 = 5.43 × 10
–27
 
kgm
s
-
Q p = mv Þ v = 
p
m
Þ v = 
27
27
5.43 10
1.67 10
-
-
´
´
 = 3.25 m/s
12. (b). V = 
2
e
150
l
volt, to determine the p.d. through which it was
accelerated to achieve the given de-broglie wavelength.
Then the same p.d. will retard it to rest. Thus,
 V = 
150
0.2 0.2 ´
 volt, V = 3765 V olt = 3.76 kV
13. (b). l
photon
 = 
hc
E
 and l
proton
 = 
h
2me
Þ
photon
electron
?
?
 
2m
=c
E
 Þ 
photon
electron
?
?
 µ 
1
E
Page 2


1. (a). l
a
 = 
0.101
V
 Å, V = 
0.101
0.004
V = 25.25 V , V = 637.5 V
E
a
 =q
a
 × V
a
 » 1275 eV
2. (d). l = 
h
mv
Q  v = 
8
c 3 10
20 20
´
=
 = 1.5 × 10
7
 m/sec
h = 6.626 × 10
–34
 J-s, m = 1.67 × 10
–27
 kg
\  
34
277
6.626 10
1.67 10 1.5 10
-
-
´
l=
´ ´´
Þ  l = 2.64 × 10
–14
 m
3. (a). Q l µ 
1
m
  Þ l
e
 µ 
e
1
m
, l
p
 µ 
p
1
m
\ 
p
e
pe
m
m
l
=
l
4. (c).  l = 
h
mv
 Þ v = 
h
ml
 ,
34
31 10
6.6 10
v
9.1 10 10 10
-
--
´
=
´ ´´
 = 7.2 × 10
5
 m/s
5. (b). l µ 
1
m
 ,
p
e
pe
m
1836
m1
l
==
l
6. (a). l
p
 = 
pp
h
2meV
Þ l
d
 = 
dd
h
2meV
\  
pp
d
p dd
me
me
l
=
l
   Q  m
d
 = 2m
p
,
e
d
 = e
p
  Þ  
pp
d
p pp
me
1
2me 2
l
==
l
7. (d). K
max
 of photoelectrons doesn't depends upon intensity
of incident light.
8. (b).
  
34
31 19
6.62 10
2 9.1 10 13.6 1.6 10
-
--
´
l=
´´ ´ ´´
Þ  l = 3.3 × 10
–10
 m = 3.3 Å
9. (c).  l = 
12.27
V
 Å
Þ V = 40–20 = 20 V olt
Þ  l = 
12.27
20
 Å = 2.75 Å
10. (d). Wavelength of electrons is l = 
150
V
Å
Now, electrons have energy of 40 KeV , therefore they
are accelerated through a potential difference of 40 ×
10
3
 volt.
       l =  
3
150
40 10 ´
 = 0.061 Å
\  Resolving limit of electron microscope = 0.061 Å
11. (d). The linear momentum of the photon
= 
34
9
h 6.63 10
122 10
-
-
´
=
l
´
 = 5.43 × 10
–27
 
kgm
s
-
Q p = mv Þ v = 
p
m
Þ v = 
27
27
5.43 10
1.67 10
-
-
´
´
 = 3.25 m/s
12. (b). V = 
2
e
150
l
volt, to determine the p.d. through which it was
accelerated to achieve the given de-broglie wavelength.
Then the same p.d. will retard it to rest. Thus,
 V = 
150
0.2 0.2 ´
 volt, V = 3765 V olt = 3.76 kV
13. (b). l
photon
 = 
hc
E
 and l
proton
 = 
h
2me
Þ
photon
electron
?
?
 
2m
=c
E
 Þ 
photon
electron
?
?
 µ 
1
E
DPP/ P 54
152
14. (d). hv – W
0
 =
2
max
1
2
mv Þ 
2
max
0
1
2
hc hc
mv -=
ll
Þ hc
2 0
max
0
1
2
mv
æö l -l
=
ç÷
ll
èø
Þ 
0
max
0
2hv
v
m
æö l -l
=
ç÷
ll
èø
When wavelength is l and velocity is v, then
0
0
2hv
v
m
æö l -l
=
ç÷
ll
èø
....(i)
When wavelength is 
3
4
l
 and velocity is 'v' then
( )
( )
0
0
3 /4 2
3 /4
hc
v
m
éù l -l
¢=
êú
l ´l
êú
ëû
...(ii)
Divide equation (ii) by (i), we get
( )
0
0
0
0
3 /4
3
4
v
v
l -l éù
ll ¢
ëû
=´
l -l
ll
( )
1/2
0
0
34
4
3
vv
l - l- éù
æöëû
¢=
ç÷
ll èø
1/2
4
.
3
ie vv
æö
¢>
ç÷
èø
15. (d). De-Broglie wavelength 
h
p
l=
Þ 
1
p
lµ
i.e. graph will be a rectangular hyperbola.
16. (d). If the incident light be of threshold wavelength (l
0
),
then the stopping potential shall be zero. Thus
l
0
 = 
hc
f
,   l
0
 = 
348
19
6.6 10 3 10
4.2 1.6 10
-
-
´ ´´
´´
l
0
 = 2.946 × 10
–7
 m  = 2946 Å
17. (b). Relation between V
0
 – n., V
0
 = 
0
h h
ee
n n
-
Put it in the form of  y = mx – c,
here V
0
 = y, n = x, 
0
h
e
n
 = c
\ y = 
h
e
æö
ç÷
èø
x – c
\ m = 
h
e
18. (b). P = 10 × 10
3
 watt,  n = ?,  l = 300 m
P  =  
nhc
t l
10
4
 = 
348
6.62 10 3 10 n
3001
-
´ ´´´
´
n = 
4
348
300 10
6.62 10 3 10
-
´
´ ´´
  = 1.5 × 10
31
19. (b). V
0
 = 
hc
el
 –
0
e
f
= 3.74 – 1.07 = 2.67 V
20. (a). The stopping potential for curves a and b is same.
\ f
a
 = f
b
Also saturation current is proportional to intensity
\ I
a
 < I
b
21. (c). hn = hn
0
 + E
k
6.6 × 10
-34
 × 3 × 10
15
 = 4 × 1.6 × 10
–19
 + E
k
19.8 × 10
-19
 – 6.4 × 10
-19
 = E
k
E
k
 = 13.4 × 10
–19 
J
Þ  
1
2
 mv
2
max
 = 13.4 × 10
–19
v
max
 = 
19
2 13.4 10
m
-
´´
    =  
19
31
2 13.4 10
9 10
-
-
´´
´
  = 1.73 × 10
6
 m/s
22. (b). The maximum kinetic energy is
K
max 
= 
hc
l
 – f = 
1242 eV nm
280 nm
-
 – 2.5 eV
         = 4.4 eV – 2.5 eV = 1.9 eV
Stopping potential V is given by eV = K
max
V = 
max
K
e
 = 
1.9
e
 eV = 1.9 V
23. (a). Q 2 d sin f = nl
l
max
 = 
max
min
(2dsin) 2d sin 90º
n1
f
=
= 2 × 10 Å
l
max
 = 20Å
\ Possible wavelengths are 5Å, 10Å and 20Å.
24. (c). l
min
 = 
12400
10000
 Å = 1.24 Å
n
max
 = 
8
10
min
c 3 10
1.24 10
-
´
=
l
´
 = 2.4×10
18
 Hz.
25. (b) DE = 
12400
4500Å
D = 2.75 eV
For photoelectric effect,  DE > W
0
 (work function).
Page 3


1. (a). l
a
 = 
0.101
V
 Å, V = 
0.101
0.004
V = 25.25 V , V = 637.5 V
E
a
 =q
a
 × V
a
 » 1275 eV
2. (d). l = 
h
mv
Q  v = 
8
c 3 10
20 20
´
=
 = 1.5 × 10
7
 m/sec
h = 6.626 × 10
–34
 J-s, m = 1.67 × 10
–27
 kg
\  
34
277
6.626 10
1.67 10 1.5 10
-
-
´
l=
´ ´´
Þ  l = 2.64 × 10
–14
 m
3. (a). Q l µ 
1
m
  Þ l
e
 µ 
e
1
m
, l
p
 µ 
p
1
m
\ 
p
e
pe
m
m
l
=
l
4. (c).  l = 
h
mv
 Þ v = 
h
ml
 ,
34
31 10
6.6 10
v
9.1 10 10 10
-
--
´
=
´ ´´
 = 7.2 × 10
5
 m/s
5. (b). l µ 
1
m
 ,
p
e
pe
m
1836
m1
l
==
l
6. (a). l
p
 = 
pp
h
2meV
Þ l
d
 = 
dd
h
2meV
\  
pp
d
p dd
me
me
l
=
l
   Q  m
d
 = 2m
p
,
e
d
 = e
p
  Þ  
pp
d
p pp
me
1
2me 2
l
==
l
7. (d). K
max
 of photoelectrons doesn't depends upon intensity
of incident light.
8. (b).
  
34
31 19
6.62 10
2 9.1 10 13.6 1.6 10
-
--
´
l=
´´ ´ ´´
Þ  l = 3.3 × 10
–10
 m = 3.3 Å
9. (c).  l = 
12.27
V
 Å
Þ V = 40–20 = 20 V olt
Þ  l = 
12.27
20
 Å = 2.75 Å
10. (d). Wavelength of electrons is l = 
150
V
Å
Now, electrons have energy of 40 KeV , therefore they
are accelerated through a potential difference of 40 ×
10
3
 volt.
       l =  
3
150
40 10 ´
 = 0.061 Å
\  Resolving limit of electron microscope = 0.061 Å
11. (d). The linear momentum of the photon
= 
34
9
h 6.63 10
122 10
-
-
´
=
l
´
 = 5.43 × 10
–27
 
kgm
s
-
Q p = mv Þ v = 
p
m
Þ v = 
27
27
5.43 10
1.67 10
-
-
´
´
 = 3.25 m/s
12. (b). V = 
2
e
150
l
volt, to determine the p.d. through which it was
accelerated to achieve the given de-broglie wavelength.
Then the same p.d. will retard it to rest. Thus,
 V = 
150
0.2 0.2 ´
 volt, V = 3765 V olt = 3.76 kV
13. (b). l
photon
 = 
hc
E
 and l
proton
 = 
h
2me
Þ
photon
electron
?
?
 
2m
=c
E
 Þ 
photon
electron
?
?
 µ 
1
E
DPP/ P 54
152
14. (d). hv – W
0
 =
2
max
1
2
mv Þ 
2
max
0
1
2
hc hc
mv -=
ll
Þ hc
2 0
max
0
1
2
mv
æö l -l
=
ç÷
ll
èø
Þ 
0
max
0
2hv
v
m
æö l -l
=
ç÷
ll
èø
When wavelength is l and velocity is v, then
0
0
2hv
v
m
æö l -l
=
ç÷
ll
èø
....(i)
When wavelength is 
3
4
l
 and velocity is 'v' then
( )
( )
0
0
3 /4 2
3 /4
hc
v
m
éù l -l
¢=
êú
l ´l
êú
ëû
...(ii)
Divide equation (ii) by (i), we get
( )
0
0
0
0
3 /4
3
4
v
v
l -l éù
ll ¢
ëû
=´
l -l
ll
( )
1/2
0
0
34
4
3
vv
l - l- éù
æöëû
¢=
ç÷
ll èø
1/2
4
.
3
ie vv
æö
¢>
ç÷
èø
15. (d). De-Broglie wavelength 
h
p
l=
Þ 
1
p
lµ
i.e. graph will be a rectangular hyperbola.
16. (d). If the incident light be of threshold wavelength (l
0
),
then the stopping potential shall be zero. Thus
l
0
 = 
hc
f
,   l
0
 = 
348
19
6.6 10 3 10
4.2 1.6 10
-
-
´ ´´
´´
l
0
 = 2.946 × 10
–7
 m  = 2946 Å
17. (b). Relation between V
0
 – n., V
0
 = 
0
h h
ee
n n
-
Put it in the form of  y = mx – c,
here V
0
 = y, n = x, 
0
h
e
n
 = c
\ y = 
h
e
æö
ç÷
èø
x – c
\ m = 
h
e
18. (b). P = 10 × 10
3
 watt,  n = ?,  l = 300 m
P  =  
nhc
t l
10
4
 = 
348
6.62 10 3 10 n
3001
-
´ ´´´
´
n = 
4
348
300 10
6.62 10 3 10
-
´
´ ´´
  = 1.5 × 10
31
19. (b). V
0
 = 
hc
el
 –
0
e
f
= 3.74 – 1.07 = 2.67 V
20. (a). The stopping potential for curves a and b is same.
\ f
a
 = f
b
Also saturation current is proportional to intensity
\ I
a
 < I
b
21. (c). hn = hn
0
 + E
k
6.6 × 10
-34
 × 3 × 10
15
 = 4 × 1.6 × 10
–19
 + E
k
19.8 × 10
-19
 – 6.4 × 10
-19
 = E
k
E
k
 = 13.4 × 10
–19 
J
Þ  
1
2
 mv
2
max
 = 13.4 × 10
–19
v
max
 = 
19
2 13.4 10
m
-
´´
    =  
19
31
2 13.4 10
9 10
-
-
´´
´
  = 1.73 × 10
6
 m/s
22. (b). The maximum kinetic energy is
K
max 
= 
hc
l
 – f = 
1242 eV nm
280 nm
-
 – 2.5 eV
         = 4.4 eV – 2.5 eV = 1.9 eV
Stopping potential V is given by eV = K
max
V = 
max
K
e
 = 
1.9
e
 eV = 1.9 V
23. (a). Q 2 d sin f = nl
l
max
 = 
max
min
(2dsin) 2d sin 90º
n1
f
=
= 2 × 10 Å
l
max
 = 20Å
\ Possible wavelengths are 5Å, 10Å and 20Å.
24. (c). l
min
 = 
12400
10000
 Å = 1.24 Å
n
max
 = 
8
10
min
c 3 10
1.24 10
-
´
=
l
´
 = 2.4×10
18
 Hz.
25. (b) DE = 
12400
4500Å
D = 2.75 eV
For photoelectric effect,  DE > W
0
 (work function).
DPP/ P 54
153
26. (a) DE = W
0
 + E  ;  (E
k
) = DE – W
0
For maximum value of (E
k
), W
0
 should be minimum
      W
0
 for lithium = 2.3 eV
\   (E
k
) = 2.75 – 2.3 = 0.45 eV
27. (c) The maximum magnitude of stopping potential will be
for metal of least work function.
\ required stopping potential is
V
s
 = 
0
hv – 
e
f
 = 0.45 volt.
28. (c) Mass of moving photon 
2
hvh
m
c
c
==
l
 and E = mc
2
29. (c) Less work function means less energy is required for
ejecting out the electrons.
30. (a) de-Broglie wavelength associated with gas molecules
varies as 
1
T
lµ