Correct answer Carries: 4.
Wrong Answer Carries: -1.
An alpha-particle with 4.0 MeV kinetic energy approaches a gold nucleus (Z = 79). What is the distance of closest approach? (Use \( \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \), \( e = 1.6 \times 10^{-19} \, \text{C} \), 1 MeV = \( 1.6 \times 10^{-13} \, \text{J} \))
\( d = \frac{2Ze^2}{4\pi\epsilon_0 K} \).
\( K = 4.0 \times 1.6 \times 10^{-13} = 6.4 \times 10^{-13} \, \text{J} \).
\( d = \frac{2 \times 79 \times (1.6 \times 10^{-19})^2 \times 9 \times 10^9}{6.4 \times 10^{-13}} \).
\( d = \frac{3.641 \times 10^{-28}}{6.4 \times 10^{-13}} \approx 5.69 \times 10^{-14} \, \text{m} \approx 57 \, \text{fm} \).
A hydrogen atom absorbs a photon of energy 1.89 eV from the \( n = 2 \) state. To which energy level does it jump? (Use \( E_n = -\frac{13.6}{n^2} \, \text{eV} \))
\( E_2 = -3.4 \, \text{eV} \).
\( E_n = -3.4 + 1.89 = -1.51 \, \text{eV} \).
\( -1.51 = -\frac{13.6}{n^2} \Rightarrow n^2 = 9 \Rightarrow n = 3 \).
In Bohr’s model, what is the physical basis for the quantization of angular momentum?
Bohr’s second postulate states that the angular momentum of the electron is an integral multiple of \( h/2\pi \), introducing quantization to ensure stable orbits.
What does de Broglie’s hypothesis explain in the context of Bohr’s model?
De Broglie’s hypothesis explains the quantization of angular momentum by proposing that electrons form standing waves, with the orbit circumference being an integral multiple of the wavelength.
Why does the classical electromagnetic theory predict that an atom should collapse?
An accelerating electron in orbit emits radiation, losing energy and spiraling into the nucleus, leading to collapse according to classical theory.
According to Bohr’s model, what happens when an electron in a hydrogen atom transitions from a higher orbit to a lower orbit?
Bohr’s third postulate states that a photon is emitted with energy equal to the difference between the initial and final energy states.
A hydrogen atom absorbs a photon of energy 12.75 eV from the ground state. To which energy level does it jump? (Use \( E_n = -\frac{13.6}{n^2} \, \text{eV} \))
\( E_1 = -13.6 \, \text{eV} \).
\( E_n = -13.6 + 12.75 = -0.85 \, \text{eV} \).
\( -0.85 = -\frac{13.6}{n^2} \Rightarrow n^2 = 16 \Rightarrow n = 4 \).
An alpha-particle with kinetic energy 6.0 MeV approaches a gold nucleus (Z = 79). What is the distance of closest approach? (Use \( \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \), \( e = 1.6 \times 10^{-19} \, \text{C} \), 1 MeV = \( 1.6 \times 10^{-13} \, \text{J} \))
\( K = 6.0 \times 1.6 \times 10^{-13} = 9.6 \times 10^{-13} \, \text{J} \).
\( d = \frac{2 \times 79 \times (1.6 \times 10^{-19})^2 \times 9 \times 10^9}{9.6 \times 10^{-13}} \).
\( d = \frac{3.641 \times 10^{-28}}{9.6 \times 10^{-13}} \approx 3.79 \times 10^{-14} \, \text{m} \approx 38 \, \text{fm} \).
A 13.0 eV electron beam excites a hydrogen atom in the ground state. What is the highest energy level reached? (Use \( E_n = -\frac{13.6}{n^2} \, \text{eV} \))
\( E_n = -13.6 + 13.0 = -0.6 \, \text{eV} \).
\( -0.6 = -\frac{13.6}{n^2} \Rightarrow n^2 \approx 22.67 \Rightarrow n = 4 \) (since \( E_4 = -0.85 \, \text{eV} < -0.6 \, \text{eV} \)).
According to Thomson’s model of the atom, how is the positive charge distributed?
In Thomson’s model, the positive charge is uniformly distributed throughout the volume of the atom, with electrons embedded in it like seeds in a watermelon.
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