(i)De Broglie wavelength of the neutron, λ = 1.40 x 10⁻¹⁰ m
Mass of a neutron, m_n = 1.66 x 10⁻²⁷ kg
Planck’s constant, h = 6.6 x 10⁻³⁴ Js
Kinetic energy (K) and velocity (v) are related as:
\(K=\frac{1}{2} m_{n} v^{2} \ldots(1)\)
De Broglie wavelength (λ) and velocity (v) are related as:
\(\lambda=\frac{h}{m_{n}} v \ldots(2)\)
Using equation (2) in equation (1), we get:
\(K=\frac{1}{2} \frac{m_{n} h^{2}}{\lambda^{2} m_{n}^{2}}=\frac{h^{2}}{2 \lambda^{2} m_{n}}\)
\(=\frac{\left(6.63 \times 10^{-34}\right)^{2}}{2 \times\left(1.40 \times 10^{-10}\right)^{2} \times 1.66 \times 10^{-27}}=6.75 \times 10^{-21} J\)
Hence, the kinetic energy of the neutron is 6.75 x 10⁻²¹ J or 4.219 x 10⁻²¹ eV.
Hence, the kinetic energy of the neutron is 6.75 x 10⁻²¹ J or 4.219 x 10⁻² eV.
(ii) Temperature of the neutron, T = 300 K
Boltzmann constant, k = 1.38 x 10⁻²³ kg m² s⁻² K⁻¹
Average kinetic energy of the neutron:
\(K \prime=\frac{3}{2} k T\)
\(=\frac{3}{2} \times 1.38 \times 10^{-23} \times 300=6.21 \times 10^{-21} J\)
The relation for the de Broglie wavelength is given as:
\(\lambda \prime=\frac{h}{\sqrt{2 K / m_{n}}}\)
Where
mn = 1.66 x 10^-27Kg
h = 6.6 x 10^-34Js
\(K \prime=6.75 \times 10^{-21} J\)
\(\therefore \lambda \prime=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 6.21 \times 10^{-21} \times 1.66 \times 10^{-27}}}=1.46 \times 10^{-10} \mathrm{~m}=0.146 \mathrm{nm}\)
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.

(i) Neha's brother was proud and happy on his sister's presence of mind, awareness of traffic rules with high visual sensibility.
(ii) The principle of photoelectric effect is used to maintain the traffic signal.
(iii) The leading physical quantity of photoelectric current is arisen due to the motion of photoelectrons.
The required equation is \(v=\sqrt { \frac { 2E }{ m } -\frac { 2\phi }{ m } } \)
where, v = speed of the electron, E = energy of the photon, m = mass of the electron and \(\phi \) = work function for the metal.

(i) This is because the energy of visible photons is less than work function of most of the metals.
(ii) \(\frac { 1 }{ 2 } { { mv }^{ 2 } }_{ max }=hv-{ { \phi }_{ 0 }=hv-{ hv }_{ 0 } }\)
 \({ v }_{ 0 }=v-\frac { { { mv }^{ 2 } }_{ max } }{ 2h } =7.21\times { 10 }^{ 14 }-\frac { \left( 9.1\times { 10 }^{ -31 }kg \right) \times { \left( 6\times { 10 }^{ 5 } \right) }^{ 2 } }{ 2\times \left( 6.63\times { 10 }^{ -34 } \right) } =4.74\times { 10 }^{ 14 }Hz\)
(iii) We find that KE of electron ejected can be increased by increasing the incident energy or by decreasing the work function. In day-to-day life we can prosper by increasing our sincere efforts or by decreasing our requirements.

6 x \({ 10 }^{ 26 }\) Na atoms weights 23 kg.
Volume of target = \({ (10 }^{ -4 }\times{ 10 }^{ -3 })={ 10 }^{ -7 }\) \({ m }^{ 3 }\)
Density of sodium = (d) = 0.97 kg/\({ m }^{ 3 }\)
Volume of \({ 6\times10 }^{ 26 }\) Na atoms = \(\frac { 23 }{ 0.97 } { m }^{ 3 }=23.7{ m }^{ 3 }\)
Volume of occupied of 1 Na atom = \(\frac { 23 }{ 0.97\times6\times{ 10 }^{ 26 } } { m }^{ 3 }\)
\(=3.95\times{ 10 }^{ -26 }=2.53\times{ 10 }^{ 18 }\)
No of sodium atoms in the target 
 \(=\frac { { 10 }^{ -7 } }{ 3.95\times{ 10 }^{ -26 } } =2.53\times{ 10 }^{ 18 }\)
Number of photons/in the beam for \({ 10 }^{ -4 }{ m }^{ 2 }=n\) Energy per second
\(=nhv={ 10 }^{ -4 }J\times100={ 10 }^{ -2 }W\)
\(hv(for\lambda =660nm)=\frac { 1234.5 }{ 600 }\)
\(=2.05eV=2.05\times1.6\times{ 10 }^{ -19 }\)
\(\\ =3.28\times{ 10 }^{ -19 }J\)
\(n\frac { { 10 }^{ -2 } }{ 3.28\times{ 10 }^{ -19 } } =3.05\times{ 10 }^{ 16/s }\)
\( n=\frac { 1 }{ 3.2 } \times{ 10 }^{ 17 }=3.1\times{ 10 }^{ 16 }\)
If p is the probability of emission per atom , per photon,the number of photoelectrons emitted/second
\(=p\times3.1\times{ 10 }^{ 16 }\times2.53\times{ 10 }^{ 18 }\)
Current \(= p\times3.1\times{ 10 }^{ +16 }\times2.53\times{ 10 }^{ 18 }\times1.6\times{ 10 }^{ -19 }A\)
This must equal \(100\mu A\)
\(p=\frac { 100\times{ 10 }^{ -6 } }{ 1.25\times{ 10 }^{ +16 } } \)
\(\therefore \)  \(p=8\times{ 10 }^{ -21 }\)

\(\Delta E=hc\left[ \frac { 1 }{ { \lambda }_{ 1 } } -\frac { 1 }{ { \lambda }_{ 2 } } \right] =\frac { hc\left( { \lambda }_{ 2 }-{ \lambda }_{ 1 } \right) }{ { \lambda }_{ 1 }{ \lambda }_{ 2 } } \)
\(\frac { 6.62\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 }\times (5000-2500)\times { 10 }^{ -10 } }{ 2500\times 5000\times { 10 }^{ -20 } } =3.96\times { 10 }^{ -19 }J\)