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 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