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In IB we learn that electrons are emitted when light is incident on a metal surface, which is a form of electromagnetic radiation.

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For every material there is a certain frequency of light under which no light is emitted.

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The work function is the energy required to take an electron from the metal. It is the energy that will release an electron such that it will have 0 kinetic energy when at the surface of the metal.

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We can use the conservation of energy to solve this question. Equating the kinetic energy with the loss in potential energy \( \frac{1}{2}mv^2 = eV \). Rearranging for \( v \) gives: \[ v = \sqrt{\frac{2Ve}{m}} \]

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Louis de Broglie proposed that all matter exhibits wave-like behavior, not just electrons or high-energy particles. This wave-particle duality applies to all particles, meaning every particle, from electrons to protons, has an associated matter wave.

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If their de Broglie wavelengths are equal, then \(\frac{h}{p_p} = \frac{h}{p_a}\). This simplifies to \(\frac{1}{m_p v_p} = \frac{1}{m_a v_a}\). Knowing that the mass of an alpha particle is four times that of a proton, its speed must be one-fourth the proton's speed to maintain the same momentum. Thus, the ratio \(\frac{v_{\text{proton}}}{v_{\text{alpha}}} = 4\).

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The correct answer is: A.

The energy of the photons is given by: \[ E = hf \] where \( h = 6.63 \times 10^{-34} \: \mathrm{Js} \), and \( f = 6.0 \times 10^{14} \: \mathrm{Hz} \). Substituting: \[ E = (6.63 \times 10^{-34})(6.0 \times 10^{14}) = 3.978 \times 10^{-19} \: \mathrm{J} \] Converting to electron volts: \[ E = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.49 \: \mathrm{eV} \] The maximum kinetic energy is given by: \[ K.E. = E - \phi = 2.49 - 2 \approx 0.5 \: \mathrm{eV} \]

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The correct answer is: C.

Electrons are emitted only if the frequency of the incident light is greater than or equal to the threshold frequency. Since \( 3.0 \times 10^{14} \: \mathrm{Hz} < 4.0 \times 10^{14} \: \mathrm{Hz} \), no electrons are emitted.

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The correct answer is: A.

The de Broglie wavelength is given by: \[ \lambda = \frac{h}{p} \] where \( h = 6.63 \times 10^{-34} \: \mathrm{Js} \), and \( p = mv = (9.1 \times 10^{-31})(2.0 \times 10^6) = 1.82 \times 10^{-24} \: \mathrm{kg \cdot m/s} \). Substituting: \[ \lambda = \frac{6.63 \times 10^{-34}}{1.82 \times 10^{-24}} \approx 3.6 \times 10^{-10} \: \mathrm{m} \]

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The correct answer is: D.

The wavelength shift is given by the Compton formula: \[ \Delta \lambda = \frac{h}{m_e c}(1 - \cos\theta) \] where \( h = 6.63 \times 10^{-34} \: \mathrm{Js} \), \( m_e = 9.11 \times 10^{-31} \: \mathrm{kg} \), \( c = 3.0 \times 10^8 \: \mathrm{m/s} \), and \( \theta = 90^\circ \). Substituting: \[ \Delta \lambda = \frac{(6.63 \times 10^{-34})}{(9.11 \times 10^{-31})(3.0 \times 10^8)}(1 - \cos90^\circ) \] \[ \Delta \lambda \approx 0.0024 \: \mathrm{nm} \]

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The correct answer is: C.

Electron diffraction, such as through a crystal lattice, demonstrates the wave nature of electrons as it produces an interference pattern, which is a property of waves.