Sodium has a work function of 2.46 eV. (a) Find the cutoff wavelength and cutoff frequency for the photoelectric effect.
(b) What is the stopping potential if the incident light has a wavelength of 181 nm?

Explanation:

given, work function of Φ = 2.46 eV.

converting the eV to joule, we have

2.45 * 1.6*10^-19 J

The cutoff wavelength is the wavelength where the incoming light does not have enough energy to free an electron, i.e. all of

the photon’s energy will be channeled into trying overcoming the work function barrier.

It is mathematically given as

Φ = hf

f = Φ/h

f = (2.46 * 1.6*10^-19) / 6.63*10^-34

f = 3.936*10^-19 / 6.63*10^-34

f = 5.94*10^14 Hz as our cut off frequency

λf = c,

λ = c/f

λ = 3*10^8 / 5.94*10^14

λ = 5.05*10^-7

λ = 505 nm as our cut off wavelength

K(max) = hf - Φ

K(max) = hc/λ - Φ

K(max) = [(6.63*10^-34 * 3*10^8) / 181*10^-9] - 3.936*10^-19

K(max) = (1.989*10^-25/181*10^-9) - 3.936*10^-19

K(max) = 1.1*10^-18 - 3.936*10^-19

K(max) = 7.064*10^-19 J or 4.415 eV

V(s) = K(max) / e

V(s) = 4.612 V

i believe the answer is a hope it and is correct