Two light sources are used in a photoelectric experiment to determine the work function for a particular metal surface.When green light from a mercury lamp (𝜆 = 546.1 nm) is used, a stopping potential of 0.8 V reduces the photocurrent to zero.Based on this measurement, what is the work function for this metal?  Express your answer in units of eV.Answer for part 1What stopping potential would be observed when using light from a red lamp with wavelength 𝜆 = 647 nm?

To solve this problem, we need to use the photoelectric effect equation:

E = hf - Φ

where E is the energy of the emitted electron, h is Planck's constant, f is the frequency of the light, and Φ is the work function of the metal.

First, we need to convert the wavelength of the green light to frequency using the equation:

f = c/λ

where c is the speed of light (3.00 x 10^8 m/s) and λ is the wavelength.

For the green light, λ = 546.1 nm = 546.1 x 10^-9 m. So,

f = (3.00 x 10^8 m/s) / (546.1 x 10^-9 m) = 5.49 x 10^14 Hz

Next, we can plug this into the photoelectric effect equation. The energy E can be found from the stopping potential V (E = eV, where e is the electron charge, 1.6 x 10^-19 C). For the green light, V = 0.8 V, so E = 1.6 x 10^-19 C * 0.8 V = 1.28 x 10^-19 J.

So,

1.28 x 10^-19 J = (6.63 x 10^-34 J*s)(5.49 x 10^14 Hz) - Φ

Solving for Φ gives Φ = 2.14 x 10^-19 J.

To convert this to eV, we divide by the charge of an electron (1.6 x 10^-19 C). So, the work function Φ = 2.14 x 10^-19 J / 1.6 x 10^-19 C = 1.34 eV.

For the second part of the question, we need to find the stopping potential for the red light. We can use the same photoelectric effect equation, but now we know the work function and need to find E.

First, we find the frequency of the red light. For the red light, λ = 647 nm = 647 x 10^-9 m. So,

f = (3.00 x 10^8 m/s) / (647 x 10^-9 m) = 4.63 x 10^14 Hz

Next, we plug this into the photoelectric effect equation:

E = (6.63 x 10^-34 J*s)(4.63 x 10^14 Hz) - (1.34 eV)(1.6 x 10^-19 J/eV)

Solving for E gives E = 1.07 x 10^-19 J.

Finally, we find the stopping potential V by dividing E by the charge of an electron: V = E/e = 1.07 x 10^-19 J / 1.6 x 10^-19 C = 0.67 V.

So, the stopping potential for the red light would be 0.67 V.