Gideon Robert University Assignment Notes

GIDEON ROBERT UNIVERSITY

SCHOOL OF MATHEMATICS AND NATURAL SCIENCES

Assignment Two

Due Date: 15/04/2026
Time: 13 hours
Lecturer: Mambwe W.

Instructions
  • Write your full name and ID number on the cover page.
  • Note that no other assignment will be given.

Problems to Solve:
  1. Calculate the radius of Bohr’s orbit in the ground state.

    • To find the radius of Bohr's orbit (specifically the first orbit or ground state), the formula is given by: rn=n2imesh24extπ2me2r_n = n^2 imes \frac{h^2}{4 ext{π}^2 m e^2} Where:
      • nn = principal quantum number (for ground state, n=1n = 1)
      • hh = Planck's constant (hext=6.626imes10−34extJ⋅sh ext{ = } 6.626 imes 10^{-34} ext{ J·s})
      • mm = mass of electron (approximately 9.11imes10−31extkg9.11 imes 10^{-31} ext{ kg})
      • ee = elementary charge (approximately 1.602imes10−19extC1.602 imes 10^{-19} ext{ C})
  2. Show that the energy of an electron in its orbit is:

    • The energy of the electron in a hydrogen atom can be expressed as:
      En=−me42extħ2n2E_n = -\frac{m e^4}{2 ext{ħ}^2 n^2}
      Where:
    • EnE_n = energy of the electron in orbit
    • mm = mass of the electron
    • ee = elementary charge
    • $ ext{ħ}$ = reduced Planck's constant ($ ext{ħ} = rac{h}{2 ext{π}}$)
    • nn = principal quantum number
  3. Calculate the longest wavelength in the Lyman series.

    • The longest wavelength in the Lyman series corresponds to the transition from n=2n = 2 to n=1n = 1.
    • Using the Rydberg formula for hydrogen:
      1extλ=RHimes(1n12−1n22)\frac{1}{ ext{λ}} = R_H imes \bigg(\frac{1}{n_1^2} - \frac{1}{n_2^2}\bigg)
      Where:
    • RH=1.097imes107extm−1R_H = 1.097 imes 10^7 ext{ m}^{-1} (Rydberg constant)
    • Solve for $ ext{λ}$ as follows:
      extλextlongest=1RHimes(112−122)ext{λ}_{ ext{longest}} = \frac{1}{R_H imes \bigg(\frac{1}{1^2} - \frac{1}{2^2}\bigg)}
  4. The longest wavelength emitted in the Paschen series.

    • The longest wavelength in the Paschen series corresponds to the transition from n=4n = 4 to n=3n = 3.
    • Applying the same Rydberg formula as in part 3:
      extλextPaschen=1RHimes(132−142)ext{λ}_{ ext{Paschen}} = \frac{1}{R_H imes \bigg(\frac{1}{3^2} - \frac{1}{4^2}\bigg)}
  5. The shortest wavelength emitted in the Balmer series.

    • The shortest wavelength in the Balmer series occurs when the transition is from n=3n = 3 to n=2n = 2.
    • Use the Rydberg formula:
      extλextshortest=1RHimes(122−132)ext{λ}_{ ext{shortest}} = \frac{1}{R_H imes \bigg(\frac{1}{2^2} - \frac{1}{3^2}\bigg)}
  6. A metal surface is receiving light of wavelength 325imes10−9extm325 imes 10^{-9} ext{ m}. If the work function for the metal is 2.46exteV2.46 ext{ eV}, calculate the maximum kinetic energy of the photoelectrons emitted from the metal surface.

    • The energy of the incident photon can be calculated using:
      E=hcextλE = \frac{hc}{ ext{λ}}
    • Where:
      • hh = Planck's constant
      • cc = speed of light (c=3.00imes108extm/sc = 3.00 imes 10^8 ext{ m/s})
    • Kinetic energy can be found using:
      K.E.=E−extWorkFunctionK.E. = E - ext{Work Function}
    • Thus:
      K.E.=hcextλ−2.46exteVK.E. = \frac{hc}{ ext{λ}} - 2.46 ext{ eV}
  7. Light of wavelength 5.00imes10−7extm5.00 imes 10^{-7} ext{ m} falls on a material that has a photoelectric work function of 2.0exteV2.0 ext{ eV}. Find:

    • (a) The energy of each individual atom.
      • The energy is given by:
        E=hcextλE = \frac{hc}{ ext{λ}}
    • (b) The kinetic energy of the most energetic photoelectron.
      • Use:
        K.E.=E−extWorkFunctionK.E. = E - ext{Work Function}
    • (c) The stopping potential.
      • The stopping potential can be calculated using:
        V=K.E.eV = \frac{K.E.}{e}
        Where ee is the charge of an electron (1.602imes10−19extC1.602 imes 10^{-19} ext{ C}).
  8. Rest mass and speed of electrons and protons.

    • The rest mass of a proton is approximately 20002000 times the rest mass of an electron.
    • To find the speed at which the electron should move so that its mass is equal to the rest mass of the proton, we use the relativistic mass increase formula:
      m=m0extsqrt(1−v2c2)m = \frac{m_0}{ ext{sqrt}(1 - \frac{v^2}{c^2})}
    • Set m=mpm = m_p (rest mass proton) and solve for vv:
      v = c imes ext{sqrt}igg{(}1 - igg{(} rac{m_e}{m_p}igg{)}^2igg{)}
    • Where:
      • mem_e is rest mass of the electron
      • mpm_p is rest mass of the proton
      • cc is the speed of light (3.00imes108extm/s3.00 imes 10^8 ext{ m/s})
  9. Radioactive Decay Process:

    • A radioactive nucleus XX undergoes β−\beta^- decay to nucleus YY.
      • (i) What could be the process giving rise to the positrons?
      • This could be a process involving β+\beta^+ decay (positron emission).
      • (ii) What are the expected end-point energies of the two positron groups?
      • These energies can be categorized as the energy levels involved in the decay process, which correspond to distinct energy values seen in the provided figure:
        • 5.53extMeV5.53 ext{ MeV}
        • 4.14extMeV4.14 ext{ MeV}
        • 1.38extMeV1.38 ext{ MeV}
        • 0extMeV0 ext{ MeV}

Conclusion
  • Ensure to follow both the provided instructions and calculations thoroughly, as accuracy is crucial for the completion of your assignment.
  • Follow up on any remaining questions or clarifications needed from your lecturer, Mambwe W.