All the electrons in the photo-sensitive material do not belong to the highest level of energy. The energies of the free electrons in the material belongs to many different closely spaced levels. So, the energies of the photoelectrons emitted from the material are distributed over a certain range.

Given, f_o = 1.25 cm, f. = - 5 cm
Magnification, m = 30 , D = 25 cm
If the object is very close to the principal focus of the objective and the image formed by the objective is very close to eyepiece, then magnifying power of a microscope is given by
\( m=-\frac{L}{f_{0}} \cdot \frac{D}{f_{e}} \)
\(\Rightarrow 30=\frac{L}{1.25} \cdot \frac{25}{5} \)
\(\Rightarrow L=\frac{125 \times 30 \times 5}{25 \times 100} \)
\(\Rightarrow L=\frac{25 \times 30}{100} \Rightarrow L=\frac{30}{4} \)
⇒  L = 7.5 cm
This is a required separation between the objective and the eyepiece.

You may think that the image will now show only half of the object, but taking the laws of reflection to be true for all points of the remaining part of the mirror, the image will be that of the whole object. However, as the area of the reflecting surface has been reduced, the intensity of the image will be low (in this case, half).

Let the resistivity of the material be \(\rho \).
∴ Resistance of wire,\(\ R=\rho \frac { l }{ A } \)
or \(\rho =\frac { RA }{ l } \)
= \(\quad \frac { 5\times 6\times { 10 }^{ -7 } }{ 15 } \)
= \(2\times { 10 }^{ -7 }\Omega -m\)
Thus the resistivity of the material at the temperature of the experiment is \(2\times { 10 }^{ -7 }\Omega -m\)

Note that thermally generated electrons (ni ~10¹⁶m^–3) are negligibly small as compared to those produced by doping.
Therefore, ne \(\approx\) ND
Since n_e n_h = \(n_{i}^{2}\) , The number of holes
n_h = (2.25 x 10³² ) / (5 x10²²)
= ~ 4.5 x 10⁹ m^–3

1u = 1.6605 x 10^–27 kg
To convert it into energy units, we multiply it by c² and find that energy equivalent = \(1.6605 \times 10^{-27} \times\left(2.9979 \times 10^{8}\right)^{2} \mathrm{~kg} \mathrm{~m}^{2} / \mathrm{s}^{2}\)
\(=1.4924 \times 10^{-10} \mathrm{~J}\)
\(=\frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-19}} \mathrm{eV}\)
\(=0.9315 \times 10^{9} \mathrm{eV}\)
\(=931.5 \mathrm{MeV}\)
or, \(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2}\)
For, \({ }_{8}^{16} \mathrm{O}, \quad \Delta M=0.13691 \mathrm{u}=0.13691 \times 931.5 \mathrm{MeV} / \mathrm{c}^{2}\)
\(=127.5 \mathrm{MeV} / \mathrm{c}^{2}\)
The energy needed to separate \({ }_{8}^{16} \mathrm{O}\) into its constituents is thus 127.5 MeV/c².

(a) \(U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r}=9 \times 10^{9} \times \frac{7 \times(-2) \times 10^{-12}}{0.18}=-0.7 \mathrm{~J}\)
(b) W = U₂ – U₁ = 0 – U = 0 – (–0.7) = 0.7 J.
(c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find
\(q_{1} V\left(\mathbf{r}_{1}\right)+q_{2} V\left(\mathbf{r}_{2}\right)=A \frac{7 \mu \mathrm{C}}{0.09 \mathrm{~m}}+A \frac{-2 \mu \mathrm{C}}{0.09 \mathrm{~m}}\)
and the net electrostatic energy is
\(q_{1} V\left(\mathbf{r}_{1}\right)+q_{2} V\left(\mathbf{r}_{2}\right)+\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r_{12}}=A \frac{7 \mu C}{0.09 m}+A \frac{-2 \mu C}{0.09 m}-0.7 \mathrm{~J}\)
= 70 − 20 − 0.7 = 49.3 J

Given, velocity of straight wire,
v = 5m/s
Horizontal component of the earth's magnetic field,

Length of the wire, \(\mathrm{I}=10 \mathrm{~m}\)
Falling speed of the wire, \(v=5.0 \mathrm{~m} / \mathrm{s}\)
Magnetic field strength, \(B=0.3 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}\)
(a) Emf induced in the wire,
\( e=B l v \)
\(=0.3 \times 10^{-4} \times 5 \times 10 =1.5 \times 10^{-3} V\)
(b) Using Fleming's right-hand rule, it can be inferred that the direction of the induced emf is from West to East.
(c) The eastern end of the wire is at a higher potential.

(a) The sizes of the atoms taken in Thomson’s model and Rutherford’s model have the same order of magnitude.
(b) In  the ground state of Thomson’s model, the electrons are in stable equilibrium. However, in Rutherford’s model, the electrons always experience a net force.
(c) A classical atom based on Rutherford’s model is doomed to collapse.
(d) An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model.
(e) The positively charged part of the atom possesses most of the mass in both the models.

Given, E₁ = 2.25 V, l₁ = 30 cm
l₂ = 60 cm, E₂ = ?
As, we know that in the case of the potentiometer, the potential gradient remains constant.
So, E ∝  l
\(\frac { E1 }{ E2 } =\frac { l1 }{ l2 } \)      .....(i)
Substituting the given values in Eq.(i), we get
\(\frac { 2.24 }{ E2 } =\frac { 30 }{ 60 } \) 
\(E2=\frac { 2.25X60 }{ 30 } =4.5\ V\)

F = Il x B
F = Il B sinθ
The force per unit length is
f = F / l = I B sinθ
(a) When the current is flowing from east to west,
θ = 90°
Hence,
f = I B
= 1 x 3x 10^–5 = 3 x 10^–5 N m^–1
This is larger than the value 2 x 10^–7 Nm^–1 quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth’s magnetic field and other stray fields while standardising the ampere.
The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors.
(b) When the current is flowing from south to north,
θ = 0^o
f = 0
Hence there is no force on the conductor.

\(\lambda =300\mu m=300\times { 10 }^{ -6 }m=3\times { 10 }^{ 8 }\)
\(E=\frac { hc }{ \lambda } =\frac { 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 300\times { 10 }^{ -6 } } J\)
\( =6.6\times { 10 }^{ -22 }J\)
\( =\frac { 6.6\times { 10 }^{ -22 } }{ 4.2 } cal=\frac { 6.6\times { 10 }^{ -22 } }{ 60\times 60 } watt \ hour\)
\( =\frac { 6.6\times { 10 }^{ -22 } }{ 1.6\times { 10 }^{ -19 } } electron \ volt\)

(a) We know that P = I V cos\(\phi \) where cos\(\phi \) is the power factor. To supply a given power at a given voltage, if cos\(\phi \) is small, we have to increase current accordingly. But this will lead to large power loss (I²R) in transmission.
(b) Suppose in a circuit, current I lags the voltage by an angle \(\phi \). Then power factor \(\phi \) = R/Z.
We can improve the power factor (tending to 1) by making Z tend to R. Let us understand, with the help of a phasor diagram.

how this can be achieved. Let us resolve I into two components. I_p along the applied voltage V and I_q perpendicular to the applied voltage. I_q as you have learnt in Section 7.7, is called the wattless component since corresponding to this component of current, there is no power loss. I_P is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit.
It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current I_q by an equal leading wattless current I'_q. This can be done by connecting a capacitor of appropriate value in parallel so that I_q and I′_q cancel each other and P is effectively I_p V.

(i) Power = +2 dioptre.
(ii) Here, we have f = +12 cm, R₁ = +10 cm, R₂ = -15 cm.
Refractive index of air is taken as unity.
We use the lens formula. The sign convention has to be applied for f, R₁ and R₂.
Substituting the values, we have
\(\frac { 1 }{ 12 } =(n-1)\left( \frac { 1 }{ 10 } -\frac { 1 }{ 15 } \right) \)
This gives n = 1.5.
(iii) For a glass lens in air, n₂ = 1.5, n₁ = 1, f = +20 cm. Hence, the lens formula gives
\(\frac { 1 }{ 20 } =0.5\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
For the same glass lens in water, n₂ = 1.5, n₁ = 1.33. Therefore \(\frac { 1.33 }{ f } =(1.5-1.33)\left[ \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right] \)
Combining these two equations, we find f = + 78.2 cm.

\(V=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2} \times \frac{4 \times 10^{-7} \mathrm{C}}{0.09 \mathrm{~m}}\)
= 4 x 10⁴ V 
(b) W = qV = 2 x 10⁻⁹C x 4 x 10⁴ V
= 8 x 10^–5 J
No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along r and another perpendicular to r. The work done corresponding to the later will be zero.

(i) (a): Angle of deviation is minimum for the red colour.
(ii) (a): In vacuum all colours have same speed, because there is no dispersion of light in vacuum.
(iii) (c): The deviation is maximum when angle is 90°.
(iv) (a): \(A=6^{\circ} ; \mu=1.644\)
\(f=(\mu-1) A \)
\(f=(1.644-1) 6=0.644 \times 6 \)
\(\delta=3.864^{\circ}\)
(v) (d): \(i_{1}=50^{\circ} ; A=60^{\circ}, \delta_{m}=?\)
\(A+\delta_{m}=i_{1}+i_{2}=50^{\circ}+50^{\circ}=100^{\circ} \)
\(\delta_{m}=100^{\circ}-A=100-60^{\circ}=40^{\circ}\)