In a photoelectric effect experiment, the threshold wavelength of the light is 380 nm. If the wavelength of incident light is 260 nm, what is the maximum kinetic energy of the emitted electrons? Given E(in eV) = 1237/λ(in nm)

The maximum kinetic energy of the emitted electrons (Kmax) in the photoelectric effect is given by:

Kmax = hν - φ

where h is Planck's constant, ν is the frequency of the incident light, and φ is the work function of the metal.

The work function can be expressed in terms of the threshold wavelength (λ0):

φ = hc/λ0

where c is the speed of light.

The energy of the incident light can be expressed in terms of its wavelength (λ):

E = hc/λ

Substituting these expressions into the equation for Kmax, we get:

Kmax = hc/λ - hc/λ0 = hc(1/λ - 1/λ0)

Given that E(in eV) = 1237/λ(in nm), we can rewrite the equation for Kmax as:

Kmax = 1237(1/λ - 1/λ0) eV

Substituting the given values, λ = 260 nm and λ0 = 380 nm:

Kmax = 1237(1/260 - 1/380) eV

Kmax = 1237(0.003846 - 0.002632) eV

Kmax = 1237(0.001214) eV

Kmax ≈ 1.5 eV