Atoms Chapter-Wise Test 3

Most of the atom is empty space, with the nucleus occupying a very small volume, so most alpha-particles do not encounter the nucleus and pass through undeflected.

The impact parameter determines the scattering angle; a smaller impact parameter leads to a larger deflection due to closer approach to the nucleus.

\( E_1 = -13.6 \, \text{eV} \).

\( E_n = -13.6 + 12.0 = -1.6 \, \text{eV} \).

\( -1.6 = -\frac{13.6}{n^2} \Rightarrow n^2 \approx 8.5 \Rightarrow n = 2 \) (since \( E_2 = -3.4 \, \text{eV} < -1.6 \, \text{eV} \)).

\( r_n = n^2 r_1 \).

\( r_5 = 25 r_1 \), \( r_1 = r_1 \).

Ratio = \( \frac{r_5}{r_1} = 25 \).

\( v_n = \frac{v_1}{n} \).

For \( n = 4 \): \( v_4 = \frac{2.2 \times 10^6}{4} = 5.5 \times 10^5 \, \text{m/s} \).

\( E_5 = -0.544 \, \text{eV} \), \( E_4 = -0.85 \, \text{eV} \).

\( \Delta E = -0.544 - (-0.85) = 0.306 \, \text{eV} \approx 0.31 \, \text{eV} \).

\( r_n = n^2 r_1 \).

For \( n = 5 \): \( r_5 = 5^2 \times 5.3 \times 10^{-11} = 25 \times 5.3 \times 10^{-11} = 1.325 \times 10^{-9} \, \text{m} \).

De Broglie’s hypothesis supports quantization by proposing standing waves, not continuous waves, which fit an integer number of wavelengths into the orbit.

Total energy \( E = K + U \), where \( K = \frac{e^2}{8\pi\epsilon_0 r} \), \( U = -\frac{e^2}{4\pi\epsilon_0 r} \).

\( E = K - 2K = -K \).

\( E = -13.6 \, \text{eV} = -K \Rightarrow K = 13.6 \, \text{eV} \).

Bohr’s model applies only to hydrogenic (single-electron) atoms and cannot account for electron-electron interactions in multi-electron atoms like helium.