(i) The de-Broglie wavelength of a particle is given as
\(\lambda=\frac{b}{\sqrt{2 m V_0 q}}\)
Since, \(\alpha\)-particle and proton both are accelerated through the same potential V₀.
\(\begin{aligned} \therefore \quad \lambda \propto \frac{1}{\sqrt{m q}} \end{aligned}\)
 \(\begin{aligned} \text { or } \quad \frac{\lambda_a}{\lambda_p}=\sqrt{\frac{m_p q_p}{m_a q_a}} \end{aligned}\)
As, charge on \(\alpha\)-particle = 2 \(\times\)charge on proton
\(q_{\mathrm{\alpha}}=2 q_{p} \Rightarrow \frac{q_{\mathrm{p}}}{q_{\mathrm{\alpha}}}=\frac{1}{2}\)
Mass of \(\alpha\)-particle = 4 \(\times\) Mass of proton
m_a = 4 \(\times\) m_p
\(\Rightarrow \quad \frac{m_p}{m_\alpha}=\frac{1}{4}\)
\(\begin{array}{ll} \therefore & \frac{\lambda_{\mathrm{a}}}{\lambda_p}=\sqrt{\frac{1}{4} \cdot \frac{1}{2}}=\frac{1}{2 \sqrt{2}}\end{array}\)
\(\begin{array}{ll} \Rightarrow & \lambda_p=2 \sqrt{2 \lambda_\alpha} \end{array}\)
i.e. proton has greater de-Broglie wavelength than that \(\alpha\)-paricle.
(ii) \(\mathrm{KE} \propto q\) (for same accelerating potential)
Since, charge on an \(\alpha\)-paricle is more as compared to a proton, so it will have a greater value of KE. Hence, proton will have lesser KE.

de-Broglie wavelength is given by
\(\lambda =\frac { h }{ \sqrt { 2mK } } =\frac { h }{ \sqrt { 2mqV } }\)
\(\lambda \propto \frac { 1 }{ \sqrt { mqV } } \)
m = mass of charge particle, q = charge and V = potential difference
Raio of de-Broglie wavelengths of proton and a-particle is given by
\(\frac { \lambda _{ \alpha } }{ \lambda _{ p } } =\sqrt { \frac { m_{ \alpha }q_{ \alpha }V_{ \alpha } }{ m_{ p }q_{ p }V_{ p } } } =\sqrt { \left( \frac { m_{ \alpha } }{ m_{ p } } \right) \left( \frac { q_{ \alpha } }{ q_{ p } } \right) \left( \frac { V_{ \alpha } }{ V_{ p } } \right) } \)
\(\frac { m_{ \alpha } }{ m_{ p } } =4,\frac { q_{ \alpha } }{ q_{ p } } =2,\frac { V_{ \alpha } }{ V_{ p } } =\frac { 1 }{ 2 } \)
a-particle is 4 toimes heavier than proton and it has double the charge than that of proton
\(\frac { \lambda _{ p } }{ \lambda _{ \alpha } } =\sqrt { 4\times 2\times \frac { 1 }{ 2 } } =2\quad \Rightarrow \lambda _{ p }:\lambda _{ \alpha }=2:1\)

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,
\(\lambda =\frac { 12.27 }{ \sqrt { V_{ } } } \mathring { A } =\frac { 12.27 }{ \sqrt { 120 } } \mathring { A=0.112\times 10^{ -9 } } \ m=0.112 \ nm\)

\( { E }_{ v }={ \phi }_{ 0 }+{ K }_{ max }\)
\( \Rightarrow { \phi }_{ 0 }=0\)
\( \Rightarrow \ { E }_{ v }={ K }_{ max }\)
\(\Rightarrow { K }_{ max }=\frac { { p }^{ 2 } }{ 2m } ={ E }_{ v }\)
\( \Rightarrow \ p=\sqrt { 2m{ E }_{ v } } \)
Wavelength \(\left( \lambda \right) \) of emitted electrons,
 \(\lambda =\frac { h }{ p } =\frac { h }{ \sqrt { 2m{ E }_{ v } } } \)