Dual Nature of Radiation and Matter Practice Questions - DPP for JEE

 Page 1


1. (a) For one photocathode
....(i)
For another photo cathode
....(ii)
Subtracting (ii) from (i) we get
– 
2. (b) As ? decreases, y increases and hence the speed of photoelectron
increases. The chances of photo electron to meet the anode
increases and hence photo electric current increases.
3. (c) Since, stopping potential is independent of distance hence new
stopping potential will remain unchanged i.e., new stopping
potential = V
0
.
4. (d) As P = 
?
p
 = ...(i)
?
e
? 2
 = ...(ii)
From equations (i) and (ii)
?
p
 ? ?
e
? 2
? 5. (a) The work function has no effect on photoelectric current so long as
h? > W
0
. The photoelectric current is proportional to the intensity of
Page 2


1. (a) For one photocathode
....(i)
For another photo cathode
....(ii)
Subtracting (ii) from (i) we get
– 
2. (b) As ? decreases, y increases and hence the speed of photoelectron
increases. The chances of photo electron to meet the anode
increases and hence photo electric current increases.
3. (c) Since, stopping potential is independent of distance hence new
stopping potential will remain unchanged i.e., new stopping
potential = V
0
.
4. (d) As P = 
?
p
 = ...(i)
?
e
? 2
 = ...(ii)
From equations (i) and (ii)
?
p
 ? ?
e
? 2
? 5. (a) The work function has no effect on photoelectric current so long as
h? > W
0
. The photoelectric current is proportional to the intensity of
incident light. Since there is no change in the intensity of light,
hence I
1
 = I
2
.
6. (c)
7. (d) h? – h?
0
 = E
K
, according to photoelectric equation, when ? = ?
0
, E
K
= 0.
Graph (d) represents E
K
 – ? relationship.
8. (a) According to relation, E = 
Because m
1
  < m
3
 < m
2
So for same ?, 
9. (d) W
0
 = h?
1
 – eV
1
= h ?
2
  – e V
2
eV
2
 = h(?
2
 – ?
1
) + eV
1
10. (a) As we know,
...(1)
...(2)
...(3)
Page 3


1. (a) For one photocathode
....(i)
For another photo cathode
....(ii)
Subtracting (ii) from (i) we get
– 
2. (b) As ? decreases, y increases and hence the speed of photoelectron
increases. The chances of photo electron to meet the anode
increases and hence photo electric current increases.
3. (c) Since, stopping potential is independent of distance hence new
stopping potential will remain unchanged i.e., new stopping
potential = V
0
.
4. (d) As P = 
?
p
 = ...(i)
?
e
? 2
 = ...(ii)
From equations (i) and (ii)
?
p
 ? ?
e
? 2
? 5. (a) The work function has no effect on photoelectric current so long as
h? > W
0
. The photoelectric current is proportional to the intensity of
incident light. Since there is no change in the intensity of light,
hence I
1
 = I
2
.
6. (c)
7. (d) h? – h?
0
 = E
K
, according to photoelectric equation, when ? = ?
0
, E
K
= 0.
Graph (d) represents E
K
 – ? relationship.
8. (a) According to relation, E = 
Because m
1
  < m
3
 < m
2
So for same ?, 
9. (d) W
0
 = h?
1
 – eV
1
= h ?
2
  – e V
2
eV
2
 = h(?
2
 – ?
1
) + eV
1
10. (a) As we know,
...(1)
...(2)
...(3)
Multiplying eqn. (2) by (3) and subtracting it from eqn.(1)
So, threshold wavelength,
11. (c) As we know,
12. (b) According to Einsten’s photoelectric effect, the K.E. of the
radiated electrons
K.E
max
 = E – W
mv
1
? 2
 = (1 – 0.5) eV = 0.5 eV
mv
2
? 2
 = (2.5 – 0.5) eV = 2 eV
  
13. (b) By using hv – hv
0
 = K
max
? h (v
1
 – v
0
) = K
1
..... (i)
And h(v
2
 – v
0
) = K
2
..... (ii)
14. (d) V = 3000 volt.
  
Page 4


1. (a) For one photocathode
....(i)
For another photo cathode
....(ii)
Subtracting (ii) from (i) we get
– 
2. (b) As ? decreases, y increases and hence the speed of photoelectron
increases. The chances of photo electron to meet the anode
increases and hence photo electric current increases.
3. (c) Since, stopping potential is independent of distance hence new
stopping potential will remain unchanged i.e., new stopping
potential = V
0
.
4. (d) As P = 
?
p
 = ...(i)
?
e
? 2
 = ...(ii)
From equations (i) and (ii)
?
p
 ? ?
e
? 2
? 5. (a) The work function has no effect on photoelectric current so long as
h? > W
0
. The photoelectric current is proportional to the intensity of
incident light. Since there is no change in the intensity of light,
hence I
1
 = I
2
.
6. (c)
7. (d) h? – h?
0
 = E
K
, according to photoelectric equation, when ? = ?
0
, E
K
= 0.
Graph (d) represents E
K
 – ? relationship.
8. (a) According to relation, E = 
Because m
1
  < m
3
 < m
2
So for same ?, 
9. (d) W
0
 = h?
1
 – eV
1
= h ?
2
  – e V
2
eV
2
 = h(?
2
 – ?
1
) + eV
1
10. (a) As we know,
...(1)
...(2)
...(3)
Multiplying eqn. (2) by (3) and subtracting it from eqn.(1)
So, threshold wavelength,
11. (c) As we know,
12. (b) According to Einsten’s photoelectric effect, the K.E. of the
radiated electrons
K.E
max
 = E – W
mv
1
? 2
 = (1 – 0.5) eV = 0.5 eV
mv
2
? 2
 = (2.5 – 0.5) eV = 2 eV
  
13. (b) By using hv – hv
0
 = K
max
? h (v
1
 – v
0
) = K
1
..... (i)
And h(v
2
 – v
0
) = K
2
..... (ii)
14. (d) V = 3000 volt.
  
= 32.6 × 10
6 
= 3.26 × 10
7
 m/s.
15. (d) Photoelectrons are emitted if the frequency of incident light is
greater than the threshold frequency.
16. (b) Work function f of metal = 2.28 eV
Wavelength of light ? = 500 nm = 500 × 10
–9
m
KE
max
 =  – f
KE
max
 =  – 2.82
KE
max
 = 2.48 – 2.28 = 0.2 ev
?
min
 = = 
= 
?
min
 =  × 10
–9
= 2.80 × 10
–9
 nm ? ? =   2.8 × 10
–9 
m
17. (a) Here, u = 0 ;  ; v = ? ; t = t
? 
de-Broglie wavelength, 
Rate of change of de-Broglie wavelength
Page 5


1. (a) For one photocathode
....(i)
For another photo cathode
....(ii)
Subtracting (ii) from (i) we get
– 
2. (b) As ? decreases, y increases and hence the speed of photoelectron
increases. The chances of photo electron to meet the anode
increases and hence photo electric current increases.
3. (c) Since, stopping potential is independent of distance hence new
stopping potential will remain unchanged i.e., new stopping
potential = V
0
.
4. (d) As P = 
?
p
 = ...(i)
?
e
? 2
 = ...(ii)
From equations (i) and (ii)
?
p
 ? ?
e
? 2
? 5. (a) The work function has no effect on photoelectric current so long as
h? > W
0
. The photoelectric current is proportional to the intensity of
incident light. Since there is no change in the intensity of light,
hence I
1
 = I
2
.
6. (c)
7. (d) h? – h?
0
 = E
K
, according to photoelectric equation, when ? = ?
0
, E
K
= 0.
Graph (d) represents E
K
 – ? relationship.
8. (a) According to relation, E = 
Because m
1
  < m
3
 < m
2
So for same ?, 
9. (d) W
0
 = h?
1
 – eV
1
= h ?
2
  – e V
2
eV
2
 = h(?
2
 – ?
1
) + eV
1
10. (a) As we know,
...(1)
...(2)
...(3)
Multiplying eqn. (2) by (3) and subtracting it from eqn.(1)
So, threshold wavelength,
11. (c) As we know,
12. (b) According to Einsten’s photoelectric effect, the K.E. of the
radiated electrons
K.E
max
 = E – W
mv
1
? 2
 = (1 – 0.5) eV = 0.5 eV
mv
2
? 2
 = (2.5 – 0.5) eV = 2 eV
  
13. (b) By using hv – hv
0
 = K
max
? h (v
1
 – v
0
) = K
1
..... (i)
And h(v
2
 – v
0
) = K
2
..... (ii)
14. (d) V = 3000 volt.
  
= 32.6 × 10
6 
= 3.26 × 10
7
 m/s.
15. (d) Photoelectrons are emitted if the frequency of incident light is
greater than the threshold frequency.
16. (b) Work function f of metal = 2.28 eV
Wavelength of light ? = 500 nm = 500 × 10
–9
m
KE
max
 =  – f
KE
max
 =  – 2.82
KE
max
 = 2.48 – 2.28 = 0.2 ev
?
min
 = = 
= 
?
min
 =  × 10
–9
= 2.80 × 10
–9
 nm ? ? =   2.8 × 10
–9 
m
17. (a) Here, u = 0 ;  ; v = ? ; t = t
? 
de-Broglie wavelength, 
Rate of change of de-Broglie wavelength
18. (a) Retarding potential depends on the frequency of incident radiation
but is independent of intensity.
19. (a) From formula
? = 
= 
[By placing value of h,  m and k)
= 
20. (b)
21. (1.5 × 10
20
) Give that, only 25% of 200W converter electrical energy
into light of yellow colour
Where N is  the No. of photons emitted per second, h = plank’s constant,
c, speed of light.
= 1.5 × 10
20
22. (1.6 × 10
15
Hz) n ? 2 – 1
E = 10.2 eV
K.E. = E – f
Q = 10.20 – 3.57
h ?
0
 = 6.63 eV