equal

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

The above curve shows that 
(i) when a heavy nucleus breaks into two medium sized nuclei(in nuclear fission)the BE/nucleon increases resulting in the release of energy
(ii) When two small nuclei combine to form relatively bigger nuclear fusion BE/nucleon increases ,resulting in the release of energy.  

Einstein's photoelectric equation,
K.E. of photoelectron = Incident energy of photons - Work function
or  K.E = hv - W₀
or  K.E = hv - hv₀
where v₀ is called threshold frequency
Threshold Frequency : For a given metal, there exists a certain minimum frequency of the incident radiation below which no emission of photoelectrons takes place. This frequency is called threshold frequency.

\(N=\frac { { N }_{ 0 } }{ { 16 }^{ ' } } \)
Where 30 years
\(N={ N }_{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ n }\)
\(\frac { N }{ { N }_{ 0 } } =\left( \frac { 1 }{ 2 } \right) ^{ 4 }\)
No.of half lives = 4
\(4=\frac { Time \ of \ disintegration }{ half \ life \ period } \)
\(\Rightarrow \frac { 30 \ years }{ 4 } =half \ life \ period\)
Half-life period = 7.5 years

Given, Kinetic energy = KE = 120 eV
p=\(\sqrt { 2eVm } =\sqrt {2KE.m }\)  \([\because K E=e V]\)
\(P=\sqrt { 2\times 120\times 1.6\times 10^{ -19 }\times 9.1\times 10^{ -31 } } \)
\(=5.91\times 10^{ -24 }\ kg-m/s\)
(ii) We know that momentum, p = mv
or, \(v=\frac{p}{m}=\frac{5.91 \times 10^{-24}}{9.1 \times 10^{-31}}\)
\(=6.5 \times 10^{6} \mathrm{~m} / \mathrm{s}\)
(iii) de-Broglie wavelength associated with electron,
\(\lambda =\frac { 12.27 }{ \sqrt { V_{ } } } \mathring { A } =\frac { 12.27 }{ \sqrt { 120 } } \mathring { A=0.112\times 10^{ -9 } } \ m=0.112 \ nm\)

Give, power = 1 W, Zener breakdown, V_Z = 5 V
Minimum voltage, V_min = 3 V
Maximum voltage, V_max = 7 V
Current, \({ I }_{ Zmax }=\frac { P }{ { V }_{ Z } } =\frac { 1 }{ 5 } =0.2A\)
The values of R_S for safe operation,
\({ R }_{ S }=\frac { { V }_{ max }-{ V }_{ Z } }{ { I }_{ Zmax } } =\frac { 7-5 }{ 0.2 } =\frac { 2 }{ 0.2 } =10\Omega\)

Metal A has higher value of work function because the slopes of both materials are constant and the intercept of the line depends on the work function.

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

Balmer series lies in the visible region.

For the emission of photoelectrons from the cesium metal, the minimum energy of the incident light on the metal surface should have the photon of energy 2eV.

The work function of a metal is the minimum energy required by an electron to just escape from the metal surface so as to overcome the restraining forces at the surface.
The relation between work function \(\left(\phi_0\right)\) and threshold frequency \(\left(v_0\right) \text { is } \phi_0=h v_0\) where h is Plank's constant. 

Let a sample of radioactive material have N₀ nuclei, at t = 0
At time t, Number of nuclei = N
As per the decay law
\(-\frac{dN}{dt}=\lambda N\)
\(\Rightarrow \ \int _{ { N }_{ 0 } }^{ N }{ \frac { dN }{ N } } =\int _{ 0 }^{ t }{ -\lambda } dt\)
\(\Rightarrow \ { \left( \log _{ e }{ N } \right) }_{ { N }_{ 0 } }^{ N }=-\left( { t }_{ 0 } \right) { t }^{ 1/2 }\)
\(\frac{N}{N_0}=e^-\lambda^ t\)
\(\Rightarrow \ \ N={ N }_{ 0 }{ e }^{ -\lambda t }\)
After one half life, Number of nuclei becomes  \(\frac{N_0}{2}\)
\(\Rightarrow \ \frac { N_{ 0 } }{ 2 } =N_{ 0 }e^{ - }\lambda ^{ T }1/2\)
\(\Rightarrow\)  2 = \(e^\lambda \)^T1/2
\(\Rightarrow\)  log_e 2 = \({ \lambda T }_{ \frac { 1 }{ 2 } }\)
\(\Rightarrow\)  \({ \lambda T }_{ \frac { 1 }{ 2 } }\)  = 0.6931
\(\Rightarrow\)  \({ T }_{ \frac { 1 }{ 2 } }\) = \(\frac{0.6931}{\lambda}\)

(a) Photo diode is fabricated with a transparent window to allow light to fall on the diode.
(b) (i) Working: When reverse biased photo diode is illuminated with light of energy greater than the forbidden energy gap (Eg), electron hole pair are generated in, or near, the depletion region. Due to junction field, electrons are collected on the n-side and holes on p-side, giving rise to a potential difference.

(a) metals, (b) insulators and (c) semiconductors
Two distinguishing features:
(i) In conductors, the valence band and conduction band tend to overlap (or nearly overlap) while in insulators they are separated by a large energy gap and in semiconductors they are separated by a smaJ1'ep.ergygap.  
(ii) The conduction band, of a conductor, has a large number of electrons available for electrical conduction. However the conduction band of insulators is almost empty while that of the semiconductor has only a (very) small number of such electrons available for electrical conduction.

(i) Heavy doping makes the depletion region very thin. This makes the electric field of the junction.
(ii) When operated under reverse bias, the photodiode can detect changes in current with changes in light intensity more easily.
(iii) The photon energy, of visible light photons varies from about 1.8 eV. Hence for visible LED's, the semiconductor must have a band gap of 1.8 eV.

(i) Three experimentally observed features in the phenomenon of photoelectric effect are as follows:
(a) Intensity When intensity of incident light increase as one [hoton ejects one electron, the increase in intensity will increase the number of ejected electrons. Frequency has no number of photoelectrons.
(b) Frequency When the frequency of incident photon increase, the knietic energy of the emitted electrons increase. Intensity has no effect on kinetic energy of photoelectron.
(c) No time lag When energy of incident photon is greater than the work function, the photoelectron is immediately ejected. Thus, there is no tile lag between the incident oflight and emission of photoelectron.
(ii) These features cannot be explained by the wave theory  of light because wave nature of  radition cannot explain the following:
(a) The instataneous ejection of photoelectrons.
(b) The existance of threshold frequency for a metal surface.
(c) The fact that kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency.

Here  V₀ = 0.60V
 \(\phi _{ 0 }=2.14\quad eV=2.14\times 1.6\times { 10 }^{ -19 }J\)
(i) Threshold frequency, \(V_{ 0 }=\frac { \phi _{ 0 } }{ h } \ [\because \ \phi _{ 0 }=hV_{ 0 }]\)]                           
 \(=\frac { 2.14\times 1.6\times { 10 }^{ -19 } }{ 6.63\times { 10 }^{ -34 } } \)     
\(=5.16\times 10^{ 14 } \ Hz\)
(ii) \(eV_{ 0 }=\frac { hc }{ \lambda } -\phi _{ 0 } \ or \ \lambda =\frac { hc }{ (eV_{ 0 }+\phi _{ 0 }) } \)
\(=\frac { 6.63\times 10^{ -34 }\times 3\times 10^{ 8 } }{ (1.6\times { 10 }^{ -19 }\times 0.60+2.14\times 1.6\times 10^{ -19 }) }\)
\( \approx 454\times { 10 }^{ -9 } \ m\)
\( =454 \ nm\)

It gives an estimate of the size of nucleus.
(ii) K.E of the ∝-particle = potential energy Possessed by beam at distance of closest approach.
\(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { (2e)(Ze) }{ { r }_{ 0 } } \)
7.7 x 1.6 x 10^-3 = \(\frac { 9\times { 10 }^{ 9 }\times 2\times 2.56\times { 10 }^{ -38 } }{ { r }_{ 0 } } \)
r₀ = \(\frac { 9\times { 10 }^{ 9 }\times 2\times 2.56\times { 10 }^{ -38 } }{ 7.7\times 1.6\times { 10 }^{ -13 } } m\)
= 299 x 10^-16 m

(a) (i) Electron moving in a circular orbit around the nucleus would get accelerated, therefore it would spiral into the nucleus, as it looses its energy.
(ii) It must emit a continuous spectrum. According to Bohr's model of hydrogen atom
(i) Electron in an atom can revolve in certain stable orbits without the emission of radiant energy

(a) Definition
(i) Half life: Time taken by a radioactive nuclei to reduce to half of the initial number of radio nuclei.
(ii)Average life: Ratio of total life time of all radioactive nuclei to the total number of nuclei in the sample.
Relation between half life and decay constant:
\({ T }_{ 1/2 }=\frac { 0.693 }{ \lambda } \)
Relation between average life and decay constant
\(\tau =\frac { 1 }{ \lambda } \)
(b) \(N={ N }_{ 0 }{ e }^{ -\lambda t }\)
\(\frac { 3 }{ 4 } { N }_{ 0 }={ N }_{ 0 }{ e }^{ -(0.3465)^{ t }\quad }\) = (\(​​​\because N\)75% of N₀)                      
\(N=\frac { 3 }{ 4 } { N }_{ 0 }\)
\({ e }^{ (0.3465)t }=\frac { 4 }{ 3 } \)
\(0.3465\times t={ log }_{ e }(4/3)\)
\(=2.303[log4-log3]\)
\(=2.303[0.6020-0.4771]\)
\( =2.303\times 0.1249\)
\( t=\frac { 2.303\times 0.1249 }{ 0.3465 } \)
\( \therefore \ t=0.83days\ or\ 19.92\ hours\)