Physics Ncert Solutions Class 12th

14.7 Energy band gap of the given photodiode, $E_g$ = 2.8 eV

Wavelength, $\lambda$ = 6000 nm = 6000 $*{10}^{-9}$ m

The energy of a signal is given by the relation,

$E$ = $\frac{hc}{\lambda }$

Where

h = Planck's constant = 6.626 $*{10}^{-34}$ Js

c = sped of light = 3 $*{10}^8$ m/s

$E$ = $\frac{6.626*{10}^{-34}*3*{10}^8}{6000*{10}^{-9}}$ = 3.313 $*{10}^{-20}$ J = $\frac{3.313*{10}^{-20}}{1.6*{10}^{-19}}$ eV = 0.207 eV

Therefore the energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV – the energy band gap of the photodiode. Hence, the photodiode cannot detect the signal.

14.6 Input frequency = 50 Hz

For a half-wave rectifier, the output frequency is equal to the input frequency.

Hence, the output frequency = 50 Hz

For a full-wave rectifier, the output frequency is twice the input frequency.

Hence the output frequency = 2 $*50Hz=100Hz$

14.5 The correct statement is (c).

When a forward bias is applied to a p-n junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced.

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)