Question What is the maximum wavelength of incident light for which photoelectrons will be released from gallium

Answer: 292 nm Explanation: The work function of gallium ∅ = 94.25 eV = 6.81 x 10^-19 J at maximum wavelength, the energy of the photons is equal to its work function Energy of the electron = hf but hf = hc/λ where h is the planck’s constant = 6.63 × 10-34 m^2 kg/s c is the speed of light = 3 x 10^8 m/s λ is the wavelength that this occurs, which is the maximum wavelength Equating, we have hc/λ = ∅ substituting, we have (6.63 × 10-34 x 3 x 10^8)/λ = 6.81 x 10^-19 (1.989 x 10^-25)/(6.81 x 10^-19) = λ λ = 292.07 x 10^-9 = 292 nm Log in to Reply

Answer:292 nmExplanation:The work function of gallium ∅ = 94.25 eV = 6.81 x 10^-19 J

at maximum wavelength, the energy of the photons is equal to its work function

Energy of the electron = hf

but hf = hc/λ

where h is the planck’s constant = 6.63 × 10-34 m^2 kg/s

c is the speed of light = 3 x 10^8 m/s

λ is the wavelength that this occurs, which is the maximum wavelength

Equating, we have

hc/λ = ∅

substituting, we have

(6.63 × 10-34 x 3 x 10^8)/λ = 6.81 x 10^-19

(1.989 x 10^-25)/(6.81 x 10^-19) = λ

λ = 292.07 x 10^-9 =

292 nm