de-Broglie wavelength is given by
\(\lambda=\frac{h}{\sqrt{2 m K}}=\frac{h}{\sqrt{2 m q V}} \quad[\because K=q V]\)
\(\Rightarrow \ \lambda \propto \frac{1}{\sqrt{m q V}}\)
Ratio of de-Broglie wavelengths of proton and a-particle is given by
\(\frac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\frac{m_{\alpha} q_{\alpha} V_{\alpha}}{m_{p} q_{p} V_{p}}}=\sqrt{\left(\frac{m_{\alpha}}{m_{p}}\right)\left(\frac{q_{\alpha}}{q_{p}}\right)\left(\frac{V_{\alpha}}{V_{p}}\right)}\)
Here, \(\frac{m_{\alpha}}{m_{p}}=4, \frac{q_{\alpha}}{q_{p}}=2\)
\(\Rightarrow \ \frac{V_{\alpha}}{V_{p}}=\frac{64}{128}=\frac{1}{2}\)
[\(\because\) a-particle is 4 times heavier than proton and it has double the charge than that of proton]
\(\Rightarrow \ \frac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{4 \times 2 \times \frac{1}{2}}=2\)
\(\Rightarrow \ \lambda_{p}: \lambda_{\alpha}=2: 1\)

\( \lambda =\frac { h }{ p } =\frac { h }{ \sqrt { 2mqV } }\)
\(As \ { m }_{ \alpha }=4{ m }_{ p } \ and \ { q }_{ \alpha }=2{ q }_{ p }\)
\(\frac { { \lambda }_{ \alpha } }{ { \lambda }_{ p } } =\sqrt { \frac { { 2m }_{ p }{ q }_{ p } }{ { 2m }_{ \alpha }{ q }_{ \alpha } } } =\sqrt { \frac { 1 }{ 8 } } =\frac { 1 }{ 2\sqrt { 2 } } 1\)

\(\lambda =\frac { h }{ mv }\)
\(=\frac { 6.63\times { 10 }^{ -34 } }{ 9.1\times { 10 }^{ -31 }\times 2.2\times { 10 }^{ 8 } }\)
\(=3.31\times { 10 }^{ -12 }m\)

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
Stopping potential : It is that minimum negative potential given to anode in a photo-cell for which the photo-electric current becomes zero. It is denoted by V_0. It is independent of the intensity of the incident light.

(i) The de-Broglie wavelength of a particle is given by
 \(\lambda =\frac { 12.27 }{ \sqrt { V_{ } } } \mathring { A } \)
[Where, V is the accelerating potential of the particle]
\(\because \quad \lambda _{ p }=\lambda _{ \alpha }\)                              [given]
\(=\frac { 12.27 }{ \sqrt { V_{ p } } } =\frac { 12.27 }{ \sqrt { V_{ \alpha } } }\)
\( \Rightarrow \frac { V_{ p } }{ V_{ \alpha } } =1\)
(ii) The de-Broglie wavelength of the particle is given by
\(\lambda =\frac { h }{ mv } \)
\( \lambda _{ p }=\frac { h }{ m_{ p }.v_{ p } } \ and \ \lambda _{ \alpha }=\frac { h }{ m_{ \alpha }v_{ \alpha } } \)
We know that, \(\\ m_{ \alpha }=4 m _{ p }\)
\( \because \lambda_{p}=\lambda_{\alpha} \)                            [given]
\( \therefore \frac{h}{m_{p} \cdot v_{p}}=\frac{h}{4 m_{p} \cdot v_{\alpha}} \Rightarrow \frac{v_{p}}{v_{\alpha}}=4 \)

Let m and K be the mass and kinetic energy of a particle. Its de-Broglie wavelength is
\(\lambda =\frac { h }{ \sqrt { 2mK } } or{ \lambda }^{ 2 }=\frac { { h }^{ 2 } }{ 2mK } or \ K=\frac { { h }^{ 2 } }{ 2m{ \lambda }^{ 2 } } \)
\( i.e.,K\alpha \frac { 1 }{ m } \)
\(\frac { { K }_{ e } }{ { K }_{ p } } =\frac { { m }_{ e } }{ { m }_{ p } } >1 or { K }_{ e }>{ K }_{ p }\)

de-Broglie wavelength,
\(\lambda =\frac { h }{ mv } =\frac { h\sqrt { 1-{ v }^{ 2 }/{ c }^{ 2 } } }{ { m }_{ 0 }v } \)
\(=\frac { h\sqrt { 1-{ c }^{ 2 }/{ c }^{ 2 } } }{ { m }_{ 0 }c } =0\)   \([\because v=c]\)

De-Broglie wavelength associated with a body of mass m, moving with velocity v is given by.\(\lambda =\frac { h }{ mv } \).
Since the mass of the objects used in our daily life is very large, hence the de-Broglie wavelength associated with them is quite small and is not visible. Hence the wave nature of matter is not more apparent to our daily observations.

(i) Energy of a photon,
\(E=hv=\frac { hc }{ \lambda } ,i.e.,E\alpha \frac { 1 }{ \lambda }\)
\(\frac { E' }{ E } =\frac { \lambda }{ 2\lambda } =\frac { 1 }{ 2 } or \ E'=\frac { E }{ 2 } \)
Thus the energy of a photon becomes half when the wavelength of radiation is doubled.
(ii) Momentum of photon,
\(p=\frac { hv }{ c } =\frac { h }{ \lambda } i.e.,\ p\alpha \frac { 1 }{ \lambda } \)
\(\\ \frac { p' }{ p } =\frac { \lambda }{ 2\lambda } =\frac { 1 }{ 2 } or\ p'=\frac { p }{ 2 } \)
Thus the momentum of photon becomes half when the wavelength of radiation is doubled.

Max.K.E. of photoelectron, \({ K }_{ max }=hv-{ \phi }_{ 0 }\) As \({ \phi }_{ 1 }>{ \phi }_{ 2 }\) , so the max. K.E. of the emitted photoelectrons will be greater for the photosensitive material having work function \({ \phi }_{ 2 }\) .

\({ v }_{ 0 }={ \phi }_{ 0 }/h=4.2\times 1.6\times { 10 }^{ -19 }/6.6\times { 10 }^{ -34 }\)
\(=6.72\times { 10 }^{ 15 }/6.6 \ Hz\)
\(v=c/\lambda =3\times { 10 }^{ 8 }/330\times { 10 }^{ -9 }=3\times { 10 }^{ 15 }/3.3Hz\)
\( =6\times { 10 }^{ 15 }/6.6Hz\)
As \(v<{ v }_{ 0 }\)therefore no photoelectric emission will take place.

(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.