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ISC Class 12 Physics Solved Question Paper 2007

Short Answer Type

41. Calculate the equivalent energy in MeV of a unified atomic mass unit.

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42. What is meant by analogue and digital signal ? What is the symbol for a XOR gate ?

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43. Draw a labelled circuit diagram of a simple oscillator (using PNP or NPN transistor in common emitter configuration). On what factors does the frequency of the oscillator depend ?

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44. Draw energy band diagrams of conductors. semiconductors and insulators.

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Long Answer Type

45. In J.J. Thomson’s experiment for the determination of the ratio of the charge to the mass (e/m) for an electron, derive the formula in terms of E, B and V where E is the electric field intensity, B is the magnetic field intensity and V is the accelerating potential between anode and cathode.

Following figure shows the experiment arrangement of J.J. Thomson’s experiment:

Plate Py is connected to the positive and P₂ to the negative of the battery; thus a uniform electric field is created which is directed along the negative Y direction.

The electron beam is deflected towards the + ve plate P_1.The force acting on the electron is,

F = e.E,

where e is the charge of the electron and E the electric intensity.

This force is directed along the + ve Y direction.

Now, a uniform magnetic field is applied in the negative Z direction in the same region as the electric field.

The electron experiences a force,

F = B.e.v. , in the negative Y direction.

If only electric field is applied the bright spot on the screen deflects upwards and strikes the screen at Q.

If only magnetic field is applied, the spot moves downwards and strikes the screen at R. If both fields are applied simultaneously, which are mutually perpendicular to each other and if the force exerted by the two simultaneously, which are mutually perpendicular to each other and if the force exerted by the two fields on the electron are equal, then the spot remains in the undeflected position P.

i.e., under the action of the crossed electric and magnetic field, the spot remains in the undeflected position P.

When the spot remains in the undeflected position P,

force exerted by the magnetic field = force exerted by the electric field.

i.e., Bev = eE.u

= e.E

or v=E/B

In the second part of the experiment only the magnetic field is applied.

The electrons move in a circular path in a uniform magnetic field.

The force acting on the electron is F = Bev, which provides the necessary centripetal force.

That is,

$Bev space equals space mv squared over straight r semicolon where space straight r space is space the space radius space of space the space circular space path. space straight e over straight m space equals space straight nu over Br But comma space straight v space equals space straight e over straight B So comma space straight e over straight m space equals space fraction numerator straight E over denominator straight B squared straight r end fraction$

If V is the potential difference between the plates P₁ and P₂ separated by a distance d then,

$straight E space equals space straight V over straight d space and space straight e over straight m space equals space fraction numerator straight E over denominator straight B squared rd end fraction space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 2 right parenthesis thin space$

Let P.D. between cathode and anode is V volts.

Velocity acquired by the electron under P.D. V is v

$1 half space m v squared space equals space e V A l s o comma space v space equals space E over B e over m space equals space fraction numerator nu squared over denominator 2 V end fraction space equals space fraction numerator E squared over denominator 2 V B squared end fraction T h e r e f o r e comma space e over m space equals space fraction numerator E squared over denominator 2 V B squared end fraction$

Hence, the result

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46.

For Bohr’s model of the hydrogen atom, the energy of electron in its ground state is found to be-13.6 eV.

(i) Draw an energy level diagram for the hydrogen atom and mark the values of energy (in eV) at n = 2 and n = ∝.

(ii) Obtain the maximum energy of a photon emitted by the hydrogen atom in eV.

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47.

(i) On what does the minimum wavelength of the continuous x-ray spectrum depend

(ii) When the filament current is increased, the intensity of the x-ray produced increases. Why?

(iii) Why is the structure of crystals studied by x-rays ?

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