Question

A sample of hydrogen atoms is irradiated with light with wavelength 86.8 nm, and electrons are observed leaving the gas. Each hydrogen atoms were initially in its ground level.

• A. The maximum kinetic energy of photoelectrons is 0.68 eV approx.
• B. The de Broglie wavelength of photoelectrons is greater than wavelength of incident photons.
• C. The de Broglie wavelength of photoelectrons is greater than that of bound electrons in H atom
• D. The magnitude of momentum of incident photons is more than momentum of ejected photoelectrons.

Answer 1

Answer: (A), (C)

The maximum kinetic energy of photoelectrons is 0.68 eV approx., The de Broglie wavelength of photoelectrons is greater than that of bound electrons in H atom

Answer 2

Here's the step-by-step solution:

1. Energy of the incident photon:
   The wavelength of the incident light is $\lambda = 86.8 \text{ nm}$.
   The energy of the photon is $E_{photon} = \frac{hc}{\lambda}$.
   Using $hc = 1240 \text{ eV nm}$, we get:
   $E_{photon} = \frac{1240 \text{ eV nm}}{86.8 \text{ nm}} \approx 14.286 \text{ eV}$.

2. Ionization energy of hydrogen atom:
   The hydrogen atom is initially in its ground state ($n=1$). The energy of the ground state is $E_1 = -13.6 \text{ eV}$.
   The energy required to remove an electron from the ground state (ionization energy) is $\Phi = |E_1| = 13.6 \text{ eV}$.

3. Maximum kinetic energy of photoelectrons:
   According to the photoelectric effect equation, the maximum kinetic energy of the ejected electron is $KE_{max} = E_{photon} - \Phi$.
   $KE_{max} = 14.286 \text{ eV} - 13.6 \text{ eV} = 0.686 \text{ eV}$.

4. Evaluate statement 1: "The maximum kinetic energy of photoelectrons is 0.68 eV approx."
   Our calculated value $0.686 \text{ eV}$ is approximately $0.68 \text{ eV}$.
   Statement 1 is correct.

5. Momentum of incident photon:
   The momentum of the incident photon is $p_{photon} = \frac{h}{\lambda}$.
   $p_{photon} = \frac{6.626 \times 10^{-34} \text{ J s}}{86.8 \times 10^{-9} \text{ m}} \approx 7.634 \times 10^{-27} \text{ kg m/s}$.

6. Momentum and de Broglie wavelength of ejected photoelectrons:
   The maximum kinetic energy is $KE_{max} = 0.686 \text{ eV}$.
   Convert $KE_{max}$ to Joules: $KE_{max} = 0.686 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} \approx 1.099 \times 10^{-19} \text{ J}$.
   The momentum of the photoelectron is $p_{electron} = \sqrt{2 m_e KE_{max}}$.
   Using $m_e = 9.109 \times 10^{-31} \text{ kg}$:
   $p_{electron} = \sqrt{2 \times 9.109 \times 10^{-31} \text{ kg} \times 1.099 \times 10^{-19} \text{ J}}$
   $p_{electron} = \sqrt{20.01 \times 10^{-50}} \approx 4.473 \times 10^{-25} \text{ kg m/s}$.
   The de Broglie wavelength of the photoelectron is $\lambda_{electron} = \frac{h}{p_{electron}}$.
   $\lambda_{electron} = \frac{6.626 \times 10^{-34} \text{ J s}}{4.473 \times 10^{-25} \text{ kg m/s}} \approx 1.481 \times 10^{-9} \text{ m} = 1.481 \text{ nm}$.

7. Evaluate statement 2: "The de Broglie wavelength of photoelectrons is greater than wavelength of incident photons."
   $\lambda_{electron} = 1.481 \text{ nm}$, $\lambda_{photon} = 86.8 \text{ nm}$.
   $\lambda_{electron} < \lambda_{photon}$.
   Statement 2 is incorrect.

8. Momentum and de Broglie wavelength of bound electron in ground state of H atom:
   In the Bohr model, the momentum of the electron in the ground state ($n=1$) is related to the Bohr radius $r_1 = a_0 = 5.29 \times 10^{-11} \text{ m}$ by the de Broglie condition $2 \pi r_1 = 1 \times \lambda_{bound, 1}$.
   $\lambda_{bound, 1} = 2 \pi r_1 = 2 \pi \times 5.29 \times 10^{-11} \text{ m} \approx 33.24 \times 10^{-11} \text{ m} = 0.3324 \text{ nm}$.
   The momentum is $p_{bound, 1} = \frac{h}{\lambda_{bound, 1}} = \frac{6.626 \times 10^{-34} \text{ J s}}{0.3324 \times 10^{-9} \text{ m}} \approx 1.993 \times 10^{-24} \text{ kg m/s}$.

9. Evaluate statement 3: "The de Broglie wavelength of photoelectrons is greater than that of bound electrons in H atom"
   $\lambda_{electron} = 1.481 \text{ nm}$, $\lambda_{bound, 1} = 0.3324 \text{ nm}$.
   $\lambda_{electron} > \lambda_{bound, 1}$.
   Statement 3 is correct.

10. Evaluate statement 4: "The magnitude of momentum of incident photons is more than momentum of ejected photoelectrons."
    $p_{photon} = 7.634 \times 10^{-27} \text{ kg m/s}$.
    $p_{electron} = 4.473 \times 10^{-25} \text{ kg m/s}$.
    $p_{photon} < p_{electron}$.
    Statement 4 is incorrect.

The correct statements are the first and the third ones.