A container consists of a sample of hydrogen like ions all of which are in the same state. In this state, ionization energy of an ion is 36 of ionization energy of hydrogen atom. This sample is now allowed to reach its most stable configuration, thereby resulting in 10 distinct types of photons. The most energetic Paschen-line photons from this sample are made incident on a metal. The stopping potential of the ejected photoelectrons is observed to be 3.704 V. Find the work function of the target (in eV) to the nearest integer.
Question
A container consists of a sample of hydrogen like ions all of which are in the same state. In this state, ionization energy of an ion is 36 of ionization energy of hydrogen atom. This sample is now allowed to reach its most stable configuration, thereby resulting in 10 distinct types of photons. The most energetic Paschen-line photons from this sample are made incident on a metal. The stopping potential of the ejected photoelectrons is observed to be 3.704 V. Find the work function of the target (in eV) to the nearest integer.
Solution
This problem involves several steps and concepts from quantum mechanics and photoelectric effect. Here's how you can solve it:
Step 1: Determine the ionization energy of the hydrogen-like ion.
The ionization energy of a hydrogen atom is 13.6 eV. Given that the ionization energy of the ion is 36 times that of hydrogen, the ionization energy of the ion is 36 * 13.6 = 489.6 eV.
Step 2: Determine the atomic number of the ion.
The ionization energy of a hydrogen-like ion is given by the formula E = 13.6 * Z^2 eV, where Z is the atomic number of the ion. Solving for Z, we get Z = sqrt(E / 13.6) = sqrt(489.6 / 13.6) = 6. This means the ion is a carbon ion with a charge of +5.
Step 3: Determine the energy of the most energetic Paschen-line photon.
The Paschen series corresponds to electron transitions ending at the third energy level (n=3). The most energetic photon in this series corresponds to a transition from the highest energy level. Given that 10 distinct types of photons are observed, the highest energy level is n = 10 + 3 = 13. The energy of this photon is given by the formula E = 13.6 * Z^2 * (1/n1^2 - 1/n2^2) = 13.6 * 6^2 * (1/3^2 - 1/13^2) = 1.94 eV.
Step 4: Determine the work function of the metal.
The stopping potential V of the ejected photoelectrons is related to the energy of the incident photon and the work function (phi) of the metal by the equation E = eV + phi, where e is the charge of an electron (1.6 * 10^-19 C). Solving for phi, we get phi = E - eV = 1.94 eV - 1.6 * 10^-19 C * 3.704 V = 1.94 eV - 0.592 eV = 1.348 eV.
Rounding to the nearest integer, the work function of the metal is approximately 1 eV.
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