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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