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CBSESolved Test Papers
PHYSICSClass XII
Chapter : Dual Nature of Matter and Radiation
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CBSE TEST PAPER-01
CLASS - XII PHYSICS (Dual Nature of Matter and Radiation)
1. Calculate the energy associated in eV with a photon of wavelength 4000Ao ? [1]
2. Mention one physical process for the release of electron from the surface of a
metal?
[1]
3. The maximum kinetic energy of photoelectron is 2.8 eV. What is the value of
stopping potential?
[1]
4. Derive an expression for debroglie wavelength of an electron? [2]
5. Light of wavelength 2000Ao falls on an aluminum surface. In aluminum 4.2 eV are
required to remove an electron. What is the kinetic energy of (a) fastest (b) the
slowest photoelectron?
[2]
6. An electromagnetic wave of wavelength is incident on a photosensitive surface
of negligible work function. If the photoelectrons emitted form this surface have
de-broglie wavelength 1 . Prove that 2
1
2.
mc
h
[2]
7. It is difficult to remove a free electron from copper than from sodium? Why? [2]
8. The following table gives the values of work functions for a few sensitive metals.
If each of these metals is exposed to radiations of wavelength 3300nm, which of
these will not exit photoelectrons and why?
S. No. Metal Work function(eV) 1 2 3
Na K Mo
1.92 2.15 4.17
[3]
9. Define threshold wavelength for photoelectric effect? Debroglie wavelength
associated with an electron associated through a potential difference V is ? What
will be the new wavelength when the accelerating potential is increase to 4V?
[3]
10. An electron has kinetic energy equal to 100eV. Calculate (1) momentum (2) speed
(3) Debroglie wavelength of the electron.
[3]
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CBSE TEST PAPER-01
CLASS - XII PHYSICS (Dual Nature of Matter and Radiation)
[ANSWERS]
Ans1: 34 8
10
6.6 10 3 10
4000 10
hcE
19
19
19
4.95 10
4.95 10
1.6 10
E J
JE eV
Ans2: Photoelectric emission.
Ans3: 212.8
2m eVo eV
2.8Vo V
Ans4: If a beam of electrons traveling through a potential difference of V volt, the electron acquires kinetic energy.
21
2
multiply by m
m eV
2 2 2
Now λ
m meV
h
m
λ2
λ Since m, e, h are constant2
heV E
meV
h
mE
Ans5: 72000 22 10oA m
4.2o eV
(a) 2
max max
1.
2K E mV hv o
E = 3.09 eV
12.27λ oA
V
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34 82
max 7
2
max
1 6.6 10 3 104.2
2 2 10
16.2 4.2 2
2
o
hcmV
mV eV
This is the K.E of the fastest electron
(b) Zero
Ans6: . .hv K E
. .
. .
hvo K E
hcK E
1
1
Using 2 . .
2
2
h
mK E
h h
mchcm
2
1
Squaring we get
2
h
mc
Or
Ans7: since o
o
hc
Where o is the threshold wavelength
Since o oNa Cu
Work function for copper is greater and it becomes difficult to remove a freeelectron from copper.
Ans8: That material will not emit photoelectrons whose work function is greater than the energy of the incident radiation.
34 8
8
19
19
19
6.6 10 3 10
33 10
6.20 10 Joules
6.20 10
1.6 10
hcE
E
EeV
Hence work function of MO is (4.17eV) which is greater than the energy of the incident radiation (= 3.76 eV) so MO will not emit photoelectrons.
Ans9: The maximum wavelength of radiation needed to cause photoelectric emission is known as threshold wavelength.
2
1
2 =
mc
n
E = 3.76 eV
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12.27
12.27 12.27'
4 2
o
o o
AV
A AV V
' 1
2
Or
Ans10: 21100
2m eV
2 19
2 17
1100 1.6 10
2
11.6 10 (1)
2
m J
m J
2 2 17
2 2 17
Multiply by m
11.6 10
2
2 1.6 10
m
m
(1) (Momentum) 72 1.6 10P m
(2) SpeedP
m24
31
5.40 10
9.1 10
(3) Debroglie wavelengthh
mV
34
24
6.6 10
5.40 10
'2
245.40 10 /P Kgm s
65.93 10 /m s
1.23 oA
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