Class 12th

14.4 The correct statement is (c).

The diffusion of charge carriers across a junction takes place from the regions of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.

14.2 The correct statement is (d).

In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms.

14.1 The correct statement is (c ).

In an n-type silicon, the electrons are the majority carrier, while the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorous, are doped in silicon atoms.

The principle of mass-energy equivalence, given by Einstein's equation:

$E = mc^2$It states that mass and energy are interchangeable.

A small amount of mass can be converted into a large amount of energy because the speed of light $(c = 3 \times 10^8 \, \text {m/s})$ is very large and appears squared in the equation.

Significance in Nuclear Reactions:

• Helps in calculating the energy release in Nuclear Reactions
• Provides explanation for powering Nuclear Reactors and the Sun
• Explanation of Binding Energy of Nucleous

Nuclear Binding Energy is the amount of energy required to completely dismantle a nucleus into its individual protons and neutrons which is equivalent to the amount of energy to form a nucleous from its constituent nucleons (protons and neutrons).

Binding energy is calculated using Einstein's mass-energy equivalence relation:

$B.E. = \Delta m \cdot c^2$Where:

• $\Delta m$ = mass defect (in kg or amu)

• $c$ = speed of light $(3 \times 10^8 \, \text {m/s})$ (3*108m/s)

The compound microscope comprises two convex lenses - the eyepiece (lower power) and the objective (high power, short focal length). The objective forms a magnified, inverted, and real image of a small object placed just beyond its focal point. Magnifying power is M = (L/f? ) * (D/f? ), where f? is the objective's focal length, L is the tube length, and f? is the eyepiece's focal length. This image acts as the object for the eyepiece.

The formula says that for a thin lens, the focal length (f) to its refractive index (? ) and radii of curvature (R? , R? ). Let a thin lens with surfaces of radii R? (first surface) and R? (second surface). We can use the refraction formula to calculate the image formation by the first surface. Following is the formula -? /v -? /u = (? -? )/R. Then we can combine both refractions and assume a lens in air (? = 1).

For mirrors and lenses, the Cartesian sign convention is used. For lenses: For convex lenses, the focal length is positive and for concave lenses, it is negative. Distances to the right of the optical center are positive.
For mirrors: For concave mirrors, the focal length is positive and for convex, it is negative. The distances to the right of the optical center are positive. The sign convention allows for consistent calculations for formulas like mirror and lens formulas, and for ray diagrams.

12.17 Mass of a negatively charged muon, $m_{?}$ = 207 $m_e$

According to Bohr's model

Bohr radius, $r_e$ $?$ $\frac{1}{m_e}$

And, energy of a ground state electronic hydrogen atom $E_e$ $?$ $m_e$

Also, the energy of a ground state muonic hydrogen atom, $E_u$ $?$ $m_u$

We have the value of the first Bohr orbit, $r_e$ = 0.53 Å = 0.53 $*{10}^{-10}$ m

Let $r_o$ be the radius of muonic hydrogen atom

At equilibrium, we can write the relation as:

$m_e{r}_{?}$ = $m_e{r}_e$

207 $m_e{r}_{?}$ = $m_e{r}_e$

$r_{?}=\frac{r_e}{207}$ = $\frac{0.53\mathrm{}*{10}^{-10}}{207}$ =2.56 $*{10}^{-13}$ m

Hence, the value of the first Bohr radius of a muonic hydrogen atom is 2.56 $*{10}^{-13}$ m

We have $E_e$ = -13.6 eV

12.16 We never speak of quantization of orbits of planets around the Sun because the angular momentum associated with planetary motion is largely relative to the value of Planck's constant (h).

The angular momentum of the Earth in its orbit is of the order of  ${10}^{70}$ h. This leads to a very high value of quantum levels n of the order of ${10}^{70}$ . For large values of n, successive energies and angular momentum are relatively very small. Hence, the quantum levels for planetary motion are considered continuous.