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Since the emission spectra is made up of discrete lines, it shows that there are discrete energy levels in an atom.

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When comparing electron transitions, the transitions ending on the first energy level always result in the greatest change in energy.

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He shot alpha particles at a gold foil and realized some particles bounced away or got deflected, which suggested the presence of something very small and positive: the nucleus.

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An isotope is an atom with the same number of protons but a different number of neutrons.

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Explaining this is still a very prominent question in today's physics, but from experiments such as reflecting/refracting light and the photoelectric effect, we can deduce it behaves both like a particle and a wave!

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The ionization energy is the energy required to remove an electron from an atom. In an energy level diagram, that would be moving to the outermost level of 0 eV.

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The correct answer is: C.

The large-angle deflections of alpha particles indicated the presence of a dense, positively charged nucleus. Such deflections could not be explained by the "plum pudding" model of the atom, which suggested a more diffuse distribution of positive charge.

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The correct answer is: A.

The number of protons is equal to \( Z = 92 \), the number of neutrons is \( A - Z = 238 - 92 = 146 \), and the number of electrons equals the number of protons for a neutral atom, so there are 92 electrons.

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The correct answer is: C.

Discrete energy levels in atoms result in electrons emitting photons of specific frequencies during transitions. These frequencies correspond to the energy differences between levels, as given by \( E = hf \).

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The correct answer is: B.

The photon energy is the difference in energy between the levels: \[ E_{\text{photon}} = E_2 - E_3 \] Substituting \( E_2 = -\frac{13.6}{2^2} = -3.4 \: \text{eV} \) and \( E_3 = -\frac{13.6}{3^2} = -1.51 \: \text{eV} \): \[ E_{\text{photon}} = -(-3.4 - (-1.51)) = 1.89 \: \text{eV} \]

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The correct answer is: A.

Substituting \( A = 64 \) into the formula: \[ R = R_0 A^{1/3} \] \[ R = 1.2 \times 64^{1/3} \] The cube root of 64 is \( 4 \): \[ R = 1.2 \times 4 = 4.8 \: \text{fm} \]

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The correct answer is: B.

The distance of closest approach is calculated using conservation of energy: \[ \text{Initial K.E.} = \text{Electrostatic P.E. at closest approach} \] \[ \frac{1}{2}mv^2 = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r} \] Rearranging for \( r \): \[ r = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{\text{K.E.}} \] Substituting \( q_1 = 2e \), \( q_2 = 79e \), \( \text{K.E.} = 6.0 \: \mathrm{MeV} = 6.0 \times 10^6 \: \mathrm{eV} \), and \( e = 1.6 \times 10^{-19} \: \mathrm{C} \): \[ r = \frac{(8.99 \times 10^9)(2 \times 1.6 \times 10^{-19})(79 \times 1.6 \times 10^{-19})}{6.0 \times 10^6 \times 1.6 \times 10^{-19}} \] \[ r \approx 37.9 \: \mathrm{fm} \]

Detailed Answer

The correct answer is: C.

At very high energies, deviations occur due to the internal structure of the nucleus (e.g., quarks or nucleon-nucleon interactions). This leads to more frequent large-angle deflections than predicted by the Rutherford scattering model, which assumes a point-like nucleus.