please shelp me with this problem

what is the stopping potential (V) if the wavelength of the incident light is 400 nm??

i mean help ..

To do this we have to know the work function of the metal.

e V = (h c/wavelength) - work function

To find the stopping potential (V), we need to use the equation for the photoelectric effect:

eV = hf - W

Where:
e is the elementary charge (1.6 x 10^-19 C)
V is the stopping potential
h is Planck's constant (6.63 x 10^-34 J·s)
f is the frequency of the incident light
W is the work function of the material

Since we are given the wavelength of the incident light (λ = 400 nm), we can use the relationship between frequency and wavelength:

c = λf

Where:
c is the speed of light (3 x 10^8 m/s)
λ is the wavelength
f is the frequency

Rearranging this equation, we get:

f = c / λ

Plugging this value into the photoelectric effect equation, we have:

eV = hc/λ - W

To solve for V, divide both sides of the equation by e:

V = (hc/λ - W) / e

Now we can substitute the known values into the equation:

c = 3 x 10^8 m/s
λ = 400 nm = 400 x 10^-9 m
h = 6.63 x 10^-34 J·s
W (work function) depends on the material used

Let's assume the work function for the material is given as 2.0 eV. Converting this to joules:

W = 2.0 eV = 2.0 x 1.6 x 10^-19 J = 3.2 x 10^-19 J

Substituting these values into the equation, we get:

V = (6.63 x 10^-34 J·s * 3 x 10^8 m/s) / (400 x 10^-9 m) - (3.2 x 10^-19 J) / (1.6 x 10^-19 C)

Simplifying this expression will give us the stopping potential (V).