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Using Bohr's model, derive expressions for the kinetic, potential, and total energy of an electron orbiting a nucleus.

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3 parts
c

E=−e28πϵ0rE = \frac{-e^2}{8\pi\epsilon_0 r}

d(i)

2.0×10−7 m2.0 \times 10^{-7} \text{ m}

d(ii)

3.3×10−27 kg m s−13.3 \times 10^{-27} \text{ kg m s}^{-1}

Step-by-step solution

10 steps

3.

Part c· 5 steps

Using Bohr's model, an electron of mass mm and charge −e-e orbits a nucleus of charge +e+e at a distance rr.

  1. Identify the forces acting on the electron.

    The electrostatic force of attraction between the electron and the nucleus provides the centripetal force for the electron's circular motion.
    Electrostatic force (FeF_e):

    Fe=14πϵ0e⋅er2=e24πϵ0r2F_e = \frac{1}{4\pi\epsilon_0} \frac{e \cdot e}{r^2} = \frac{e^2}{4\pi\epsilon_0 r^2}

    Centripetal force (FcF_c):

    Fc=mv2rF_c = \frac{mv^2}{r}

  2. Equate the forces to find an expression for mv2mv^2.

    mv2r=e24πϵ0r2\frac{mv^2}{r} = \frac{e^2}{4\pi\epsilon_0 r^2} mv2=e24πϵ0rmv^2 = \frac{e^2}{4\pi\epsilon_0 r}

  3. Calculate the kinetic energy (KK) of the electron.

    K=12mv2K = \frac{1}{2}mv^2

    Substitute the expression for mv2mv^2:

    K=12(e24πϵ0r)=e28πϵ0rK = \frac{1}{2} \left( \frac{e^2}{4\pi\epsilon_0 r} \right) = \frac{e^2}{8\pi\epsilon_0 r}

  4. Calculate the potential energy (UU) of the electron.

    The potential energy of an electron (charge −e-e) in the electric field of a nucleus (charge +e+e) at a distance rr is:

    U=14πϵ0(+e)(−e)r=−e24πϵ0rU = \frac{1}{4\pi\epsilon_0} \frac{(+e)(-e)}{r} = \frac{-e^2}{4\pi\epsilon_0 r}

  5. Calculate the total energy (EE).

    The total energy is the sum of the kinetic and potential energies:

    E=K+UE = K + U E=e28πϵ0r+(−e24πϵ0r)E = \frac{e^2}{8\pi\epsilon_0 r} + \left( \frac{-e^2}{4\pi\epsilon_0 r} \right)

    To combine these terms, find a common denominator:

    E=e28πϵ0r−2e28πϵ0rE = \frac{e^2}{8\pi\epsilon_0 r} - \frac{2e^2}{8\pi\epsilon_0 r} E=e2−2e28πϵ0rE = \frac{e^2 - 2e^2}{8\pi\epsilon_0 r} E=−e28πϵ0rE = \frac{-e^2}{8\pi\epsilon_0 r}

    Thus, the total energy of an electron in a hydrogen atom is E=−e28πϵ0r\boxed{E = \frac{-e^2}{8\pi\epsilon_0 r}}.

Part d· 1 step

Given:
Energy of radiation, E=6.2 eVE = 6.2 \text{ eV}
Planck's constant, h=6.6×10−34 J sh = 6.6 \times 10^{-34} \text{ J s}
Speed of light, c=3.0×108 m s−1c = 3.0 \times 10^8 \text{ m s}^{-1}
Conversion factor, 1 eV=1.6×10−19 J1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}

  1. Convert the energy from electron volts (eV) to Joules (J).

    E=6.2 eV×(1.6×10−19 J/eV)E = 6.2 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV}) E=9.92×10−19 JE = 9.92 \times 10^{-19} \text{ J}

Part d(i)

2.0×10−7 m2.0 \times 10^{-7} \text{ m}

Calculate the wavelength (λ\lambda) of the emitted radiation.

  1. Use the formula relating energy, Planck's constant, speed of light, and wavelength

    E=hcλE = \frac{hc}{\lambda}.
    Rearrange to solve for λ\lambda:

    λ=hcE\lambda = \frac{hc}{E}

  2. Substitute the values and calculate λ\lambda.

    λ=(6.6×10−34 J s)×(3.0×108 m s−1)9.92×10−19 J\lambda = \frac{(6.6 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m s}^{-1})}{9.92 \times 10^{-19} \text{ J}} λ=19.8×10−269.92×10−19 m\lambda = \frac{19.8 \times 10^{-26}}{9.92 \times 10^{-19}} \text{ m} λ≈1.9959677×10−7 m\lambda \approx 1.9959677 \times 10^{-7} \text{ m}

    Rounding to two significant figures:

    λ≈2.0×10−7 m\lambda \approx \boxed{2.0 \times 10^{-7} \text{ m}}

Part d(ii)

3.3×10−27 kg m s−13.3 \times 10^{-27} \text{ kg m s}^{-1}

Calculate the momentum (pp) of the photons.

  1. Use the formula relating momentum, energy, and speed of light

    p=Ecp = \frac{E}{c}.

  2. Substitute the values and calculate pp.

    p=9.92×10−19 J3.0×108 m s−1p = \frac{9.92 \times 10^{-19} \text{ J}}{3.0 \times 10^8 \text{ m s}^{-1}} p=3.3066...×10−27 kg m s−1p = 3.3066... \times 10^{-27} \text{ kg m s}^{-1}

    Rounding to two significant figures:

    p≈3.3×10−27 kg m s−1p \approx \boxed{3.3 \times 10^{-27} \text{ kg m s}^{-1}}
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