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).