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.

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

E = hf - Φ

where E is the energy of the ejected 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 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. But first, we need to convert the wavelength from nm to m:

λ = 546.1 nm = 546.1 x 10^-9 m

Now we can find the frequency:

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

Next, we can find the energy of the ejected electron using the equation:

E = eV

where e is the charge of an electron (1.60 x 10^-19 C) and V is the stopping potential (0.8 V):

E = (1.60 x 10^-19 C)(0.8 V) = 1.28 x 10^-19 J

Now we can find the work function using the photoelectric effect equation:

Φ = hf - E

Substituting the values we found:

Φ = (6.63 x 10^-34 J*s)(5.49 x 10^14 Hz) - 1.28 x 10^-19 J = 2.36 x 10^-19 J

Finally, we need to convert the work function from J to eV. Since 1 eV = 1.60 x 10^-19 J, we have:

Φ = 2.36 x 10^-19 J * (1 eV / 1.60 x 10^-19 J) = 1.47 eV

So, the work function for this metal is 1.47 eV.