Since the emission spectra is made up of discrete lines, it shows that there are discrete energy levels in an atom.
CloseWhen comparing electron transitions, the transitions ending on the first energy level always result in the greatest change in energy.
CloseHe 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.
CloseAn isotope is an atom with the same number of protons but a different number of neutrons.
CloseExplaining 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!
CloseThe 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.
CloseThe 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.
CloseThe 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.
CloseThe 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 \( E = hf \).
CloseThe 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 \( E_3 = -\frac{13.6}{3^2} = -1.51 \: \text{eV} \): \[ E_{\text{photon}} = -(-3.4 - (-1.51)) = 1.89 \: \text{eV} \]
CloseThe 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 \( 4 \): \[ R = 1.2 \times 4 = 4.8 \: \text{fm} \]
CloseThe 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 \( e = 1.6 \times 10^{-19} \: \mathrm{C} \): \[ r = \frac{(8.99 \times 10^9)(2 \times 1.6 \times 10^{-19})(79 \times 1.6 \times 10^{-19})}{6.0 \times 10^6 \times 1.6 \times 10^{-19}} \] \[ r \approx 37.9 \: \mathrm{fm} \]
CloseThe 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.
Close| Option | Observation |
|---|---|
| A | Most alpha particles passed straight through the foil. |
| B | Alpha particles were scattered at small angles. |
| C | A few alpha particles were deflected through large angles. |
| D | Alpha particles were absorbed by the gold foil. |
| Option | Protons, Neutrons, Electrons |
|---|---|
| A | 92, 146, 92 |
| B | 92, 92, 146 |
| C | 146, 92, 92 |
| D | 92, 146, 146 |
| Option | Explanation |
|---|---|
| A | Atoms emit photons of all possible frequencies. |
| B | Atoms emit photons only when electrons move in circular orbits. |
| C | Atoms emit photons as electrons move between discrete energy levels. |
| D | Atoms emit photons when the nucleus absorbs energy. |