Physics

CBSE Class 12

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31.

(a) Draw the plot of binding energy per nucleon (BE/A) as a function of mass number A. Write two important conclusions that can be drawn regarding the nature of nuclear force.

(b) Use this graph to explain the release of energy in both the processes of nuclear fusion and fission.

(c) Write the basic nuclear process of neutron undergoing –decay. Why is the detection of neutrinos found very difficult?

Graphical representation of (BE/A) for nucleons with mass number A.

The variation of binding energy per nucleon VS. mass number is shown in the figure:

Characteristics of Nuclear force:

(i) Nuclear forces non-central and short ranged force.

(ii) Nuclear forces between proton-neutron and neutron-neutron are strong and attractive in nature.

b) When a heavy nucleus (A > 235 say) breaks into two lighter nuclei (nuclear fission), the binding energy per nucleon increases i.e, nucleons get more tightly bound. This implies that energy would be released in nuclear fission.

When two very light nuclei (A £10) join to form a heavy nucleus, the binding is energy per nucleon of fused heavier nucleus more than the binding energy per nucleon of lighter nuclei, so again energy would be released in nuclear fusion.

c) During the decay process of neutron, we have

Neutrinos show weak interaction with other particles. Hence, its detection is very different.

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(a) Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.

(b) Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero.

Electric dipole moment is the product of either charges or the distance between two equal and opposite charges.

It is a vector quantity.

Electric dipole moment at a point on the equatorial plane:

Consider a point P on broad side on the position of dipole formed of charges + q and - q at separation 2l. The distance of point P from mid-point O of electric dipole is r.

Let E_{1} and E_{2} be the electric field strength due to charges +q and –q of electric dipole.

From the fig. we have

Now, inorder to find the resultant electric field, we resolve the components along and perpendicular to AB.

The components perpendicular to AB are sin components and they being equal and opposite to each other cancel each other.

Therefore,

Resultant electric field is given by,

E_{1} = E_{1}

But,

From the fig. we can see that,

If dipole is infinitesimal and point P is far away, then *l*^{2} can be neglected as compared to r^{2}.

Therefore,

b) Equipotential surfaces due to an electric dipole is given by,

Electric potential is zero at all points in the plane passing through the dipole equator.

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33.

Using Gauss’ laws deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R at a point (i) outside and (ii) inside the shell.

Plot a graph showing variation of electric field as a function of r > R and r < R. (r being the distance from the centre of the shell).

i) Consider a uniformly charged thin spherical shell of radius R carrying charge Q. To find the electric field outside the shell, we consider a spherical Gaussian surface of radius (r >R), concentric with given shell. If E is electric field outside the shell, then by symmetry electric field strength has same magnitude E_{0} on the Gaussian surface and is directed radially outward.

So, electric flux through Gaussian surface is given by,

Therefore,

Charge enclosed by the Gaussian surface is Q.

Therefore, using gauss’s theorem, we have

Thus, electric field outside a charged thin spherical shell is the same as if the whole charge Q is concentrated at the centre.

ii) Electric field inside the shell:

The charge resides on the surface of a conductor. Thus, a hollow charged conductor is equivalent to a charged spherical shell. Let’s consider a spherical Gaussian surface of radius (r < R). If E is the electric field inside the shell, then by symmetry electric field strength has the same magnitude E_{i} on the Gaussian surface and is directed radially outward.

Electric flux through the Gaussian surface is given by,

=

Now, Gaussian surface is inside the given charged shell, so charge enclosed by Gaussian surface is zero.

Therefore, using Gauss’s theorem, we have

Thus, electric field at each point inside a charged thin spherical shell is zero.

The graph above shows the variation of electric field as a function of R.

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34.

Using Bohr’s postulates, derive the expression for the frequency of radiation emitted when electron in hydrogen atom undergoes transition from higher energy state (quantum number n_{i }) to the lower state, (n_{f} ).

When electron in hydrogen atom jumps from energy state n_{i} =4 to n_{f} =3, 2, 1, identify the spectral series to which the emission lines belong.

According to Bohr’s postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge +e. For an electron moving with a uniform speed in a circular orbit on a given radius, the centripetal force is provided by the Coulomb force of attraction between the electron and the nucleus.

Therefore,

... (1)

So, Kinetic Energy, K.E =

Potential energy is given by, P.E =

Therefore, total energy is given by, E = K.E + P.E =

E = , is the total energy.

For nth orbit, E can be written as E_{n},

... (2)

Now, using Bohr's postulate for quantization of angular momentum, we have

Putting this value of v in equation (1), we get

Now, putting value of r_{n} in equation (2), we get

R is the rydberg constant.

For hydrogen atom Z =1,

If n_{i} and n_{f }are the quantum numbers of initial and final states and E_{i} & E_{f} are energies of electron in H-atom in initial and final state, we have

That is, when electron jumps from n_{i} = 4 to n_{f} = 3.21 .

Radiation belongs to Paschen, Balmer and Lyman series.

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