Both Thomson's and Rutherford's models describe an atom of the same overall size. In Thomson's model, electrons can be in stable equilibrium inside the positive charge distribution. In Rutherford's model, orbiting electrons experience centripetal force and, classically, should radiate energy and collapse into the nucleus. Thomson's model spreads mass and positive charge nearly continuously, while Rutherford's model concentrates most mass and all positive charge in the tiny nucleus. In both models, most of the atomic mass is associated with the positively charged part.
(a) no different from. (b) Thomson’s model; Rutherford’s model. (c) Rutherford’s model. (d) Thomson’s model; Rutherford’s model. (e) both the models.
Gold nuclei are heavy and highly charged, so an alpha-particle can be strongly repelled while the nucleus remains nearly fixed. A hydrogen nucleus has charge $+e$ and mass much smaller than a gold nucleus and comparable to the alpha-particle mass. The target nucleus would recoil appreciably, and the Coulomb repulsion would be much weaker. Therefore the Rutherford experiment with solid hydrogen would not show the same appreciable large-angle scattering seen with gold foil.
Large-angle back scattering would be extremely rare or absent; most alpha-particles would pass through with small deflections.
The emitted photon has energy $E=h\nu=2.3\,\text{eV}=2.3(1.602\times10^{-19})=3.68\times10^{-19}\,\text{J}$. Hence $\nu=E/h=(3.68\times10^{-19})/(6.626\times10^{-34})=5.6\times10^{14}\,\text{Hz}$.
$5.6\times10^{14}\,\text{Hz}$.
For a Coulomb orbit in the hydrogen atom, $K=-E$ and $U=2E$, where $E=K+U$ is the total energy. With $E=-13.6\,\text{eV}$, $K=+13.6\,\text{eV}$ and $U=-27.2\,\text{eV}$.
Kinetic energy $=+13.6\,\text{eV}$; potential energy $=-27.2\,\text{eV}$.
For hydrogen, $E_n=-13.6/n^2\,\text{eV}$. Thus $E_1=-13.6\,\text{eV}$ and $E_4=-13.6/16=-0.85\,\text{eV}$. The absorbed photon energy is $\Delta E=12.75\,\text{eV}$. Therefore $\lambda=hc/\Delta E=1240/12.75=97.3\,\text{nm}$ and $\nu=\Delta E/h=12.75(1.602\times10^{-19})/(6.626\times10^{-34})=3.08\times10^{15}\,\text{Hz}$.
$\lambda=97.3\,\text{nm}$ and $\nu=3.08\times10^{15}\,\text{Hz}$.
In Bohr's model, $v_n=v_1/n$ with $v_1=2.18\times10^6\,\text{m s}^{-1}$. Hence $v_2=1.09\times10^6\,\text{m s}^{-1}$ and $v_3=7.27\times10^5\,\text{m s}^{-1}$. The orbit radius is $r_n=n^2a_0$, with $a_0=5.29\times10^{-11}\,\text{m}$. Thus $T_n=2\pi r_n/v_n=2\pi n^2a_0/(v_1/n)=n^3T_1$. Now $T_1=2\pi(5.29\times10^{-11})/(2.18\times10^6)=1.5\times10^{-16}\,\text{s}$, so $T_2=8T_1=1.2\times10^{-15}\,\text{s}$ and $T_3=27T_1=4.1\times10^{-15}\,\text{s}$.
(a) $v_1=2.18\times10^6\,\text{m s}^{-1}$, $v_2=1.09\times10^6\,\text{m s}^{-1}$, $v_3=7.27\times10^5\,\text{m s}^{-1}$. (b) $T_1=1.5\times10^{-16}\,\text{s}$, $T_2=1.2\times10^{-15}\,\text{s}$, $T_3=4.1\times10^{-15}\,\text{s}$.
Bohr radii scale as $r_n=n^2r_1$. With $r_1=5.3\times10^{-11}\,\text{m}$, $r_2=4r_1=2.12\times10^{-10}\,\text{m}$ and $r_3=9r_1=4.77\times10^{-10}\,\text{m}$.
$r_2=2.12\times10^{-10}\,\text{m}$ and $r_3=4.77\times10^{-10}\,\text{m}$.
At room temperature, hydrogen atoms are mostly in the ground state. A $12.5\,\text{eV}$ electron can excite hydrogen from $n=1$ to $n=2$ because $\Delta E_{1\to2}=10.2\,\text{eV}$, and to $n=3$ because $\Delta E_{1\to3}=13.6(1-1/9)=12.09\,\text{eV}$. It cannot excite $n=4$, which requires $12.75\,\text{eV}$. Possible emissions are $2\to1$, $3\to1$ and $3\to2$. Their energies are $10.2\,\text{eV}$, $12.09\,\text{eV}$ and $1.89\,\text{eV}$, giving wavelengths $\lambda=1240/E$: $121.6\,\text{nm}$, $102.6\,\text{nm}$ and $656.5\,\text{nm}$.
The emitted wavelengths are about $121.6\,\text{nm}$, $102.6\,\text{nm}$ and $656.5\,\text{nm}$.
Bohr quantisation gives $mvr=n h/(2\pi)$, so $n=2\pi mvr/h$. Substituting $m=6.0\times10^{24}\,\text{kg}$, $v=3.0\times10^4\,\text{m s}^{-1}$ and $r=1.5\times10^{11}\,\text{m}$ gives $mvr=2.7\times10^{40}\,\text{kg m}^2\text{s}^{-1}$. Hence $n=2\pi(2.7\times10^{40})/(6.626\times10^{-34})=2.6\times10^{74}$.
$n\approx2.6\times10^{74}$.