(i) Given, mass of bullet, m = 0.040 kg
Speed of bullet, v = 1 km/s = 1 \(\times\) 10³ m/s
de-Broglie wavelength,
\(\lambda=\frac{b}{m v}=\frac{6.63 \times 10^{-34}}{0.040 \times 1 \times 10^3}\)
= 1.66 \(\times\) 10^-35 m
(ii) Mass of the ball, m = 0.060 kg and speed of the ball, v=1 m/s
\(\begin{aligned} \lambda & =\frac{b}{m v}=\frac{6.63 \times 10^{-34}}{0.060 \times 1} \end{aligned}\)
\(\begin{aligned} =1.1 \times 10^{-32} \mathrm{~m} \end{aligned}\)
(iii) Mass of a dust particle, m = 1 \(\times\) 10^-9 kg and speed of the dust particle, v = 2.2 m/s
\(\lambda=\frac{b}{m v}=\frac{6.63 \times 10^{-34}}{1 \times 10^{-9} \times 2.2}\)
= 3.0 \(\times\) 10^-25 m

0.149 cm

For a particle, de Broglie wavelength, \(\lambda\) = h/p
Kinetic energy, K = p²/2m
Then, \(\lambda\) = h / \(\sqrt{2 m K}\)
For the same kinetic energy K, the de Broglie wavelength associated with the particle is inversely proportional to the square root of their masses. A proton \(\left(\begin{array}{l} 1 \\ 1 \end{array} \mathrm{H}\right)\) is 1836 times massive than an electron and an \(\alpha\)-particle \(\left(\begin{array}{l} 4 \\ 2 \end{array} \mathrm{He}\right)\) four times that of a proton. Hence, \(\alpha\) – particle has the shortest de Broglie wavelength.

(a) For the decay process

     \(\quad \quad _{ 88 }{ { Ra }^{ 223 } }\rightarrow _{ 82 }{ { Pb }^{ 209 } }+_{ 6 }{ { C }^{ 14 } }+Q\\ \quad \quad Q=[{ m }_{ N }(_{ 88 }{ { Ra }^{ 223 } })-{ m }_{ N }(_{ 82 }{ { Pb }^{ 209 } })-{ m }_{ N }(_{ 6 }{ { C }^{ 14 } })]{ c }^{ 2 }\\ \quad \quad \quad =[{ m }(_{ 88 }{ { Ra }^{ 223 } })-{ m }(_{ 82 }{ { Pb }^{ 209 } })-{ m }(_{ 6 }{ { C }^{ 14 } })]{ c }^{ 2 }\\ \quad \quad \quad =[223.01850-208.98107-14.00324]{ uc }^{ 2 }\\ or\quad Q=[0.03419]{ uc }^{ 2 }\\ \quad \quad \quad \quad =0.03419X931.5\quad MeV\\ \quad \quad \quad \quad =31.85\quad MeV\) 

      \(For\quad the\quad decay\quad process\\ \begin{matrix} 223 \\ 88 \end{matrix}Ra\rightarrow \begin{matrix} 219 \\ 86 \end{matrix}Rn+\begin{matrix} 4 \\ 2 \end{matrix}He+Q\\ Q=[{ m }_{ N }(\begin{matrix} 223 \\ 88 \end{matrix}Ra)-{ m }_{ N }(\begin{matrix} 219 \\ 86 \end{matrix}Rn)-{ m }_{ N }(\begin{matrix} 4 \\ 2 \end{matrix}He)]{ c }^{ 2 }\\ \quad =[{ m }_{ N }(\begin{matrix} 223 \\ 88 \end{matrix}Ra)-{ m }_{ N }(\begin{matrix} 219 \\ 86 \end{matrix}Rn)-{ m }_{ N }(\begin{matrix} 4 \\ 2 \end{matrix}He)]{ c }^{ 2 }\\ \quad =(223.01850-219.00948-4.00260){ uc }^{ 2 }\\ \quad =(0.00642){ uc }^{ 2 }\\ \quad =0.0642X931.5\quad MeV\\ \quad =5.98\quad MeV.\)

m_p = 1u, m_∝ = 4u and q_p = e, q_∝ = 4e
\(\frac{1}{2}\)mv² = qV
p = mv =\({ \left[ \sqrt { 2qVm } \right] }^{ \frac { 1 }{ 2 } }\)
\(\frac { { P }_{ p } }{ { P }_{ \alpha } } =\frac { 1 }{ 8 } \)

According to question,
P ⟶ Q + Q
BE/A of element P =7.6MeV (given)
So, BE of P = 7.6 x 240 MeV  [A = 240]
BE/A of element Q = 8.5 MeV given
So, BE of Q = 8.5 x 120 MeV
Now, energy released = 2 ( BE of Q) - BE of P
= 8.5 x 120 x 2 - 7.6 x 240
⇒  = (2040 -1824) MeV
⇒ = 216 MeV

Energy of one photon 
\(E=\frac { 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 6\times { 10 }^{ -7 } } \)  
\( \simeq 3.3\times { 10 }^{ -19 }J\)
E₁ = energy emitted by the source in one second = 66J
\(\therefore \) Number of photons emitted by the source in
\(1s=n=\frac { 66 }{ 3.3\times { 10 }^{ -19 } } =2\times { 10 }^{ 20 }\)
\(\therefore \) Total number of photons emitted by source in 2 minutes
\(=N=n\times 2\times 60\)
\( =2\times { 10 }^{ 20 }\times 120\)
\(=2.4\times { 10 }^{ 22 }photons\)

\(2.12\times { 10 }^{ -10 }\ m\ and\ 4.47\times { 10 }^{ -10 }\ m\)

A chain reaction may die out due to any of the following reasons:
(i) Size of fissionable material may be less than the critical size.
(ii) Mass of fissionable material may be less than the critical mass.
(iii) Neutron absorbing material (arrestor) might absorb neutrons at a faster rate than the rate at which they are being produced.

No, the photo-electric emission from a metal surface is possible if the frequency of the incident light is greater than the threshold frequency v_o for that metal or the energy of the incident photon is greater than the work function of the metal, \(\text { i.e., } v>v_0 \text { where. } \phi_0=h v_0\)

While drawing the plot. we have to keep in mind that first binding energy will increase sharply and then it will be constant almost.
For plot of binding energy per nucleon as the function of mass number A
Following are the two conclusions that can be drawn regarding the nature of the nuclear force.
The force is attractive and strong enough to produce a binding energy of few MeV per nucleon.
The two important conclusions regarding the nature of nuclear force are given below.
(i) The nuclear force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon.
(ii) The constancy of the binding energy in the wide range of mass number 30 < A < 170 indicate that nuclear force is a short-range force.
(b) (i) According to the binding energy curve, a very heavy nucleus (A > 170), has lower binding energy per nucleon compared to nuclei of middle mass number (30 < A < 170).
Thus, if a heavy nucleus breaks into two nuclei of mass number between 30 and 170, nucleons get more tightly bound. This implies energy would be released in the process. (nuclear fission)
(ii) When two light nuclei (A < 10) join to form a heavier nucleus, the binding energy per nucleon of fused heavier nucleus increases.
Again it indicates that energy would be released in the process (nuclear fusion).
(c) The basic nuclear process of neutron undergoing β-decay is given as
\(n \rightarrow p+e^{-}+\bar{v}\)
Here \(\bar{v}\) is antinutrino.
Neutrino and antineutrino both are neutral particles with very small (possibly, even zero) mass compared to the electrons. They have only weak interaction with other particles. Therefore, the detection of neutrinos is found very difficult.

A person approaching a doorway may interrupt a light beam which is incident on a photocell. The abrupt change in photocurrent records every interruption of the light beam caused by the person passing across the beam. In this way it helps count the person entering the auditorium, provided they enter the hall one by one.

\(Here\quad \lambda =0.82\overset { \circ }{ A } =0.82\times { 10 }^{ -10 }m\)
\( { \phi }_{ 0 }=0\)
From Einstein's photoelectric equation.
K.E of electron,
\(\frac { 1 }{ 2 } { mv }^{ 2 }=hv-{ \phi }_{ 0 }\)
\( or \ { mv }^{ 2 }=2hv=\frac { 2hc }{ \lambda } \ \ (\because { \phi }_{ 0 }=0)\)
\( or \ mv=\sqrt { \frac { 2hcm }{ \lambda } } \)
De-Broglie wavelength associated with the electron is given by
\({ \lambda }^{ ' }=\frac { h }{ mv } =\frac { h }{ \sqrt { 2hcm/\lambda } } =\sqrt { \frac { h\lambda }{ 2cm } } \)
\(=\sqrt { \frac { 6.6\times { 10 }^{ -34 }\times 0.82\times { 10 }^{ -10 } }{ 2\times 3\times { 10 }^{ 8 }\times 9.1\times { 10 }^{ -31 } } } \)
\(=0.099\times { 10 }^{ -10 }m\)
\( =0.099\overset { \circ }{ A } \)