Final answer:
To find the work function of metal B, we use the equations derived from the photoelectric effect, calculating energy balances for both metals A and B and solving for the unknown work function. The kinetic energy relations and de-Broglie wavelength ratio lead us to determine that the work function of metal Option B is 2 eV. right.
Step-by-step explanation:
The student is asking about the work function of metal B when electrons are ejected from its surface by incoming photons. The photoelectric effect is used to determine the work function, which is the minimum energy required to remove an electron from a metal surface. When a photon of energy 4.0 eV strikes metal A, the maximum kinetic energy of the ejected photoelectrons is denoted by Ta. If these photoelectrons have a de-Broglie wavelength λa, then for metal B which is hit by a 4.50 eV photon and the photoelectrons have kinetic energy (Ta + 1.5) eV and de-Broglie wavelength 2λa, we need to find the work function of metal B.
According to the photoelectric effect, the kinetic energy (KE) of an ejected photoelectron can be expressed as:
KE = hv - Φ
where hv is the energy of the incoming photon and Φ is the work function of the metal. For metal A,
4.0 eV = Ta + ΦA
For metal B with photon energy 4.50 eV and the kinetic energy (Ta + 1.5) eV:
4.50 eV = (Ta + 1.5) eV + ΦB
Subtracting the kinetic energy equation of metal A from the one for metal B gives us:
(4.50 eV - ΦB) - (4.0 eV - ΦA) = 1.5 eV
Solving for ΦB gives us the work function for metal B. Using the given de-Broglie wavelength ratio, we find that ΦB = 2 eV, which is answer option (b).