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Atomic and Nuclear Physics Practice Problems for AP Physics B

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By — McGraw-Hill Professional
Updated on Feb 12, 2011

Review the following concepts if necessary:

Problems

Multiple Choice:

  1. Which of the following lists types of electromagnetic radiation in order from least to greatest energy per photon?
    1. red, green, violet, ultraviolet, infrared
    2. ultraviolet, infrared, red, green, violet
    3. red, green, violet, infrared, ultraviolet
    4. infrared, red, green, violet, ultraviolet
    5. ultraviolet, violet, green, red, infrared
  2. Questions 2 and 3

    In a nuclear reactor, uranium fissions into krypton and barium via the reaction

  3. What are the mass number A and atomic number Z of the resulting krypton nucleus?
  4. How much mass is converted into the kinetic energy of the resulting nuclei?
    1. 1 amu
    2. 2 amu
    3. 3 amu
    4. zero
    5. much less than 1 amu
  5. decays via β+ emission. Which of the following is the resulting nucleus?

Atomic and Nuclear Physics

  1. A hypothetical atom has two energy levels, as shown above.
  2. (a) What wavelengths of electromagnetic radiation can be absorbed by this atom? Indicate which of these wavelengths, if any, represents visible light.

    (b)Now, monochromatic 180-nm ultraviolet radiation is incident on the atom, ejecting an electron from the ground state. What will be

    1. the ejected electron's kinetic energy
    2. the ejected electron's speed
    3. the incident photon's speed

    For parts (c) and (d), imagine that the 180-nm radiation ejected an electron that, instead of being in the ground state, was initially in the –1.2 eV state.

    (c) Would the speed of the ejected electron increase, decrease, or stay the same? Justify your answer briefly.

    (d) Would the speed of the incident photon increase, decrease, or stay the same? Justify your answer briefly.

Solutions

  1. D—The radiation with the highest frequency (or shortest wavelength) has the highest energy per photon by E = hf. In the visible spectrum, red has the longest wavelength and violet has the shortest. Outside the visible spectrum, infrared radiation has a longer wavelength than anything visible, and ultraviolet has a shorter wavelength than anything visible. So, infrared has the smallest energy per photon, and so on up the spectrum to ultraviolet with the most energy per photon.
  2. A—The total number of protons + neutrons is conserved. Before the reaction, we have one free neutron plus 235 protons and neutrons in the uranium, for a total of 236 amu. After the reaction, we have 141 amu in the barium plus 3 free neutrons for a total of 144 amu… leaving 92 AMUs for the krypton.
  3. Charge is also conserved. Before the reaction, we have a total charge of +92 from the protons in the uranium. After the reaction, we have 56 protons in the barium. Since a neutron carries no charge, the krypton must account for the remaining 36 protons.

  4. E—Einstein's famous equation is written ΔE = mc2, because it is only the lost mass that is converted into energy. Since we still have a total of 236 amu after the reaction, an entire amu of mass was not converted to energy. Still, the daughter particles have kinetic energy because they move. That energy came from a very small mass difference, probably about a million times less than one amu.
  5. C—In β+ emission, a positron is ejected from the nucleus. This does not change the total number of protons + neutrons, so the atomic mass A is still 15. However, charge must be conserved. The original O nucleus had a charge of +8, so the products of the decay must combine to a charge of +8. These products are a positron, charge +1, and the daughter nucleus, which must have charge +7.
  6. (a) ΔE = hc/λ, so hcE = λ. hc = 1240 eV·nm, as found on the equation sheet. ΔE represents the difference in energy levels. An electron in the ground state can make either of two jumps: it could absorb a 2.1-eV photon to get to the middle energy level, or it could absorb a 3.3 eV photon to escape the atom. An electron in the middle state could absorb a 1.2-eV photon to escape the atom. That makes three different energies. Convert these energies to wavelengths using ΔE = hc/λ, so hcE = Δ. hc = 1240 eV·nm, as found on the equation sheet; ΔE represents the energy of the absorbed photon, listed above. These photons thus have wavelengths of 590 nm for the E1 to E2 transition; 380 nm or less for the E1 to E∞ transition; and 1030 nm or less for the E2 to E∞ transition. Only the 590 nm wavelength is visible because the visible spectrum is from about 400–700 nm.
  7. (b)

    1. Find the energy of the incident photon using ΔE = hc/λ, getting 6.9 eV. This is the total energy absorbed by the electron, but 3.3 eV of this is used to escape the atom. The remaining 3.6 eV is kinetic energy.
    2. To find speed, set kinetic energy equal to 1/2mv2. However, to use this formula, the kinetic energy must be in standard units of joules. Convert 3.6 eV to joules by multiplying by 1.6 × 10-19 J/eV (this conversion is on the constant sheet), getting a kinetic energy of 5.8 × 10-19 J. Now solve the kinetic energy equation for velocity, getting a speed of 1.1 × 106 m/s. This is reasonable—fast, but not faster than the speed of light.
    3. A photon is a particle of light. Unless it is in an optically dense material (which the photons here are not), the speed of a photon is always 3.0 × 108 m/s.

    (c) The electron absorbs the same amount of energy from the incident photons. However, now it only takes 1.2 eV to get out of the atom, leaving 5.7 eV for kinetic energy. With a larger kinetic energy, the ejected electron's speed is greater than before.

    (d) The speed of the photon is still the speed of light, 3.0 × 108 m/s.

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