(i) Main use of photodiode In demodulation optical signal and detection of optical signal. In light operated switches, electronic counters.
(ii) Main use of Zener diode As DC voltage regulator.

Circuit diagram of forward biased and reverse biased p-n junction diode is shown below

The width of depletion layer
(i) decreases in forward bias.
(ii) increases in reverse bias.

The photodiode can be used as a photodetector to detect optical signals. In light operated switches.

In α-decay, the mass number of parent nucleus decreases by 4 units and atomic number decreases by 2 units.
\(\therefore 238_{U_{02}} \stackrel{\alpha-\text { decay }}{\longrightarrow} 234 T h_{90}\)

During β-decay,
(i) no change in mass number.
(ii) atomic number increases by 1.

\(T_{1 / 2}=\frac{\operatorname{In} 2}{\lambda} \tau I n 2\)

Zener diode
Alternatively


After the breakdown voltage, there is an insignificant change in the (reverse) bias voltage across the Zener diode (and hence the load) even for large changes in current.

Working:
During one-half of the input AC, the diode is forward biased and a current flows through RL. During the other half of the input AC, the diode is reverse biased and no current flows through the load RL. Hence, the given AC input is rectified.

From the V-I characteristics of a junction diode. It is clear that it allows the current to pass only when it is forward biased. So when an alternatively voltage is applied across the diode, current flows only during that part of the cycle when it is forward biased.

The SI unit of activity is becquerel
1 becquerel = 1 Bq = 1 decay/sec.

Nuclear forces are the strongest forces in nature.They are effective only inside the nucleus.

(a) Overlapping of conduction band and valence band
(b) A large energy gap (more than 3 eV) Between conduction band and valence band.

(i) Mass defect \((\triangle M)\) ,of any nucleus \(_{ Z }^{ A }{ X }\) is the sum of masses of the nucleus (=M) and the sum of masses of its constituent nucleons (=M')
\(\triangle M={ M }^{ ' }-M\)
\(=\left[ { Zm }_{ p }+(A-Z){ m }_{ n } \right] -M\)
where \({ m }_{ p }\) and \({ m }_{ n }\) denote the mass of the proton and the neutron respectively.
(ii) Binding energy is the energy required to seperate a nucleus into its constituent nucleons. The relation between the two is B.E=(mass defect) c²
(iii) There is a release of energy i.e., the reaction is exothermic. Reason: Increase in B.E/nucleon implies that more mass has been converted into energy.

(a) \(\frac { dN }{ dt } =-\lambda N\)
\(\int _{ { N }_{ 0 } }^{ N }{ \frac { dN }{ N } =\int _{ 0 }^{ t }{ -\lambda dt } } \) 
\(\left[ { log }_{ e }^{ N } \right] _{ { N }_{ 0 } }^{ N }=-\lambda \left[ t \right] _{ 0 }^{ t }\)
\(loge\frac { N }{ { N }_{ 0 } } =-\lambda t\)
\(N={ N }_{ 0 }{ e }^{ -\lambda t }\)
(b) (i) \(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }Ne+{ e }^{ x }+\upsilon \)
Also accept,if a student does not identify the product nucleus and writes as
\(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }Xe+{ e }^{ x }+\upsilon \)
Basic process  
\(p\rightarrow n+{ e }^{ + }+\upsilon \)
(ii) Isobar

Let N₀ and N be the values of the number of atoms in a radioactive sample at t=0 and at t=40 years, respectively.
Here, N = \(\frac { 1 }{ 16 } \) N₀, t = 40 years
Decay constant is given by 
\(\lambda =\frac { 2.303 }{ t } \log _{ 10 }{ \left( \frac { { N }_{ 0 } }{ N } \right) } =\frac { 2.303 }{ 40 } \log _{ 10 }{ \left( \frac { { N }_{ 0 } }{ { N }_{ 0 }/16 } \right) }\)
\(=\frac { 2.303 }{ 40 } \times \log _{ 10 }{ 16 } =\frac { 2.303 }{ 40 } \times 1.2041\)
\( \lambda =0.0693 \ { year }^{ -1 }\)

80min.

\(N={ N }_{ 0 }-\frac { 19 }{ 20 } { N }_{ 0 }=\frac { 1 }{ 20 } { N }_{ 0 }\)
\( Use \ \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ n }; \ t=nT\)

equal