This model could not explain scattering of \(\alpha\) particle through large angles.

Ionisation energy is the minimum energy required to knock out an electron from an atom.Its value will be different for different atoms. Ionisation energy will also depend on the orbit from which electron is to be removed.
When an electron in hydrogen atom is replaced by a particle 200 times heavier than electron but having the same charge, ionisation energy will not change, as it depends only on charge and not on mass of particle.

When an electron of mass m is confined to move on a line of length l with velocity v, the de-Broglie wavelength \(\lambda \) associated with electron is  \(\lambda =\frac { h }{ mv } =\frac { h }{ p } \ \ or \ \ p=\frac { h }{ \lambda } =\frac { h }{ { 2l }/{ n } } =\frac { nh }{ 2l } \)
When electron revolves in a circular orbit of radius r; then \(2l=2\pi r\)
\( \ \therefore \ \ p=\frac { nh }{ 2\pi r } \\ \ \ or \ \ p\times r=\frac { nh }{ 2\pi } \)
i.e., angular momentum  \((p\times r)\) of electron is integral multiple of \({ h }/{ 2\pi }\) . This is Bohr's quantization condition of angular momentum.

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

V₁ = 25 V, V₂ = 36 V
De-Broglie wavelength of an electron:
\( \lambda=\frac{1.227}{\sqrt{V}} \\ \lambda_1=\frac{12.27}{\sqrt{V_1}} \\ \lambda_2=\frac{12.27}{\sqrt{V_2}} \\ \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{V_2}{V_1}}=\sqrt{\frac{36}{25}} \\ \frac{\lambda_1}{\lambda_2}=\frac{6}{5} \)

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

\(\lambda =\frac { h }{ \sqrt { 2meV } } or \ \lambda =\frac { 12.27 }{ \sqrt { V } } \overset { 0 }{ A }\)
\(\lambda =\frac { 12.27 }{ \sqrt { 100 } } \overset { 0 }{ A } \ =1.227\overset { 0 }{ A } \)
This wavelength corresponds to the X-rays.

\(10944\mathring { A } \)
From  \(\frac { 1 }{ \lambda } =R\left( \frac { 1 }{ { n }_{ 1 }^{ 2 } } -\frac { 1 }{ { n }_{ 2 }^{ 2 } } \right) \)
For first member of Lyman series,
\({ n }_{ 1 }=1, \ { n }_{ 2 }=2\)
\(\frac { 1 }{ \lambda } =R\left( \frac { 1 }{ { 1 }^{ 2 } } -\frac { 1 }{ { 2 }^{ 2 } } \right) =\frac { 3 }{ 4 } R \ \therefore \ R=\frac { 4 }{ 3\lambda } \)
For 3^rd member of Paschen series,
 \({ n }_{ 1 }=3, \ { n }_{ 2 }=6\)
\(\therefore \frac { 1 }{ \lambda \prime } =R\left( \frac { 1 }{ 3^{ 2 } } -\frac { 1 }{ 6^{ 2 } } \right) =R\times \frac { 3 }{ 36 } =\frac { R }{ 12 } \)
\(\lambda \prime =\frac { 12 }{ R } =\frac { 12 }{ { 4 }/{ 3\lambda } } =9\lambda \)
\(=9\times 1216\mathring { A } =10944\mathring { A } \)

\(3646.8\mathring { A } \)
\(Take \ { n }_{ 1 }=2 \ and \ { n }_{ 2 }=\infty \)

According to photon picture:
(i) Each quantum of radiation has energy hv
(ii) In photo - electric effect the electrons in the metal absorbs this quantum of energy (hv).
(iii) When this energy exceeds the minimum energy needed for the ejection of photoelectron, flow of photo current starts.

Given, cut-off voltage, V₀ = 1.5 V
Maximum kinetic energy is given by,
\(\begin{aligned} \mathrm{KE}_{\text {max }} & =e V_0=1.5 \space \mathrm{eV}=1.5 \times 1.6 \times 10^{-19} \end{aligned}\)
\(=2.4 \times 10^{-19} \mathrm{~J}\)

Kinetic energy of the electron, E_k = 120 eV
Planck’s constant, h = 6.6 x 10⁻³⁴ Js
Mass of an electron, m = 9.1 x 10⁻³¹ kg
Charge on an electron, e = 1.6 x 10⁻¹⁹ C
(a) For the electron, we can write the relation for kinetic energy as:
\(E_{k}=\frac{1}{2} m v^{2}\)
Where,`
v = Speed of the electron
\(\therefore v^{2}=\sqrt{\frac{2 e R_{k}}{m}}\)
\(=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 120}{9.1 \times 10^{-31}}}\)
\(=\sqrt{42.198 \times 10^{12}}=6.496 \times 10^{6} \mathrm{~m} / \mathrm{s}\)
Momentum of the electron, p = mv
= 9.1 x 10⁻³¹ x 6.496 x 10⁶
= 5.91 x 10⁻²⁴ kg m s⁻¹
Therefore, the momentum of the electron is 5.91 x 10⁻²⁴ kg m s⁻¹_.
(b) Speed of the electron, v = 6.496 x 10⁶ m/s
(c) De Broglie wavelength of an electron having a momentum p, is given as:
\(\lambda=\frac{h}{p}\)
\(=\frac{6.6 \times 10^{-34}}{5.91 \times 10^{-24}}=1.116 \times 10^{-10} \mathrm{~m}\)
= 0.112 nm
Therefore, the de Broglie wavelength of the electron is 0.112 nm.

(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