In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 x 10⁻²⁷ kg) is less than the mass of incident α−particles (6.64 x 10⁻²⁷ kg). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α−particles would not bounce back if solid hydrogen is used in the α-particle scattering experiment.

The ratio of the radius of electron’s orbit to the radius of nucleus is (10^–10 m) /(10^–15 m) = 10⁵, that is, the radius of the electron’s orbit is 10⁵ times larger than the radius of nucleus. If the radius of the earth’s orbit around the sun were 10⁵ times larger than the radius of the sun, the radius of the earth’s orbit would be 10⁵ x 7 x 10⁸ m = 7 x 10¹³ m. This is more than 100 times greater than the actual orbital radius of earth. Thus, the earth would be much farther away from the sun. It implies that an atom contains a much greater fraction of empty space than our solar system does.

We have
\(R={ R }_{ 0 }{ A }^{ \frac { 1 }{ 3 } }\)
\(\therefore \)  Density \(\rho \) = \(\frac { mA }{ \frac { 4 }{ 3 } \pi \left( { R }_{ 0 }{ A }^{ \frac { 1 }{ 3 } } \right) ^{ 3 } } \)
=\(\frac { m }{ \frac { 4 }{ 3 } \pi { R }_{ 0 }^{ 3 } } \)
Hence is independent of A. (Here m is the mass of the nucleus).

When white light is passed through a gas and analysed by spectrometer, we find some dark lines in the spectrum. These dark lines correspond precisely to those wavelengths which were found in the emission line spectrum of the gas. This is called absorption spectrum.

Electron and proton have almost infinite lifetime.

As radius, \(r\propto \frac { 1 }{ m } \)
\(\therefore \) When electron is replaced by a particle 200 times heavier, the radius would decrease to \(\frac { 1 }{ 200 } \) time the original radius.

Trajectory of a-particles in the coulomb field of target nucleus is shown below:

From this experiment, the following points are observed.
(i) Most of the a-particles pass straight through the gold foil. It means that they do not suffer any collision with gold atoms.
(ii) About one a-particle in every 8000 a-particles deflects by more than 90°. As most of the a-particles go undeflected and only a few get deflected, this shows that most of the space in an atom is empty and at the centre of the atom, there exists a nucleus. By the number of a-particles get deflected, the information regarding size of the nucleus can be known.

(i) An electron undergoes transition from second excited state to the first excited state which corresponds to Balmer series and then to the ground state which corresponds to Lyman series.
(ii) The wavelength of the emitted radiations in the two cases.

We know that,\(\lambda=\frac{h c}{\Delta E}\)
Frorn n₃ \(\rightarrow\) n2,
\(\lambda_{1}=\frac{h c}{E_{3}-E_{2}}\)
\(=\frac{h c}{(-1.5)-(-3.4)}=\frac{h c}{1.9}\)
From n2 \(\rightarrow\) n₁
\(\lambda_{2}=\frac{h c}{E_{2}-E_{1}}\)
\(=\frac{h c}{(-3.4)-(-13.6)}=\frac{h c}{10.20}\)
\(\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\frac{10.20}{1.9}=5.3\)

we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10^–11 m is 2.2 × 10^–6 m/s. Thus, the frequency of the electron moving around the proton is
\(v=\frac{v}{2 \pi r}=\frac{2.2 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}}{2 \pi\left(5.3 \times 10^{-11} \mathrm{~m}\right)}\)
\(\approx \) 6.6 × 10¹⁵ Hz
According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is 6.6 × 10¹⁵ Hz.

Total energy of the electron in hydrogen atom is –13.6 eV = –13.6 × 1.6 × 10^–19 J = –2.2 ×10^–18 J. Thus from Equation we have
\(-\frac{e^{2}}{8 \pi \varepsilon_{0} r}=-2.2 \times 10^{-18} \mathrm{~J}\)
This gives the orbital radius
\(r=-\frac{e^{2}}{8 \pi \varepsilon_{0} E}=-\frac{\left(9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}^{2}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2}}{(2)\left(-2.2 \times 10^{-18} \mathrm{~J}\right)}\)
= 5.3 × 10^–11 m
The velocity of the revolving electron can be computed from Eq.uation with m = 9.1 × 10^–31 kg,
\(v=\frac{e}{\sqrt{4 \pi \varepsilon_{0} m r}}=2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

Bohr's quantization postulate is in terms of Planck's constant (h). But angular momenta associated with planetary motion are \(\approx { 10 }^{ 70 }h\) . In terms of Bohr's quantization postulate, this will correspond to \(n\approx { 10 }^{ 70 }\) For such large values of n, the differences in successive energies and angular momenta of the quantized levels are so small, that the levels can be considered as continuous and not discrete.

\(9\times { 10 }^{ 14 }J\)

Shortcomings of Rutherford's atomic model.
(i) Line spectrum
Rutherford's atomic model could not explain the sharply defined discrete wavelengths in the hydrogen spectrum.
(ii) Stability of atom
Rutherford's orbital motion of electrons around the nucleus could not explain the stability of an atom. According to electron dynamics an accelerated electron radiate energy, so the radius of orbit goes on decreasing, and finally, the electron will fall into the nucleus and the atom will not be stable.
Bohr's model: Bohr removes the drawbacks of the Rutherford model based on plank's quantum theory. According to Bohr
(1) Electron can revolve in well defined energy levels in which its angular momentum is integral multiple of \(\left(\frac{h}{2 \pi}\right)\)
\(m v r=\frac{n h}{2 \pi^{\prime}}\)    
n = principal quantam number
n = 1,2,3...
(2) Electron revolving in a definite orbit does not radiate energy, it will radiate energy only when changing the orbit. The energy of an electron in an orbit is
\(E_n=-\frac{R h C}{n^2}=\frac{-13.6}{n^2} e V\)

On the ascending part of the curve, B.E./A increases with the mass number (A).
So, B.E. of the resultant will be greater than that of the nuclei which are fused together. Hence the fusion of the nuclei on the ascending part results in an increase in binding energy, therefore, more stable nucleus
On the descending part of the curve, the B.E. of the heavier nuclei is lower. So the fission of the heavier nuclei, on the descending part, will cause an increase in B.E. and, therefore, more stable nucleus

The given nuclear reaction is:
\({ }_{6}^{11} C \rightarrow_{5}^{11} B+e^{+} v\)
Half life of \(-6^{11} C \text { nuclei } T_{\frac{1}{2}}=20.3 \mathrm{~min}\)
Atomic mass of \(m\left({ }_{6}^{11} C\right)=11.011434 \mathrm{u}\)
Atomic mass of \(m\left({ }_{6}^{11} B\right)=11.009305 u\)
Maximum energy possessed by the emitted positron = 0.960 MeV
The change in the Q-value (ΔQ) of the nuclear masses of the \({ }_{6}^{11} C\) nucleus is given as:
\(\left.\triangle Q=\left[m \prime{(6} C^{11}\right)-\left[m{ }^{\prime}\left({ }_{5}^{11} B\right)+m_{e}\right]\right] c^{2} \ldots .(1)\)
Where,
m_e = Mass of an electron or positron = 0.000548 u
c = Speed of light
m’ = Respective nuclear masses
If atomic masses are used instead of nuclear masses, then we have to add 6 m_e in the case of ¹¹C and 5 m_e in the case of ¹¹B
Hence, equation (1) reduces to:
\(\triangle Q=\left[m\left({ }_{6} C^{11}\right)-m\left({ }_{5}^{11} B\right)-2 m\right] c^{2} \text { are the atomic masses }\)
\(\text { Here } m\left({ }_{6} C^{11}\right) \text { and } m\left({ }_{5}^{11} B\right) \text { are the atomic masses }\)
∴ΔQ = [11.011434 − 11.009305 − 2 x 0.000548] c²
= (0.001033 c²) u
But 1 u = 931.5 Mev/c²
∴ΔQ = 0.001033 x 931.5 ≈ 0.962 MeV
The value of Q is almost comparable to the maximum energy of the emitted positron.