Light of wavelength 350 nm falls on a potassium surface, and the photoelectrons have a maximum kinetic energy of 1.3 eV.
What is the work function of potassium? The speed of light is 3 × 108 m/s and Planck’s constant is 6.63×10−34J·s.
What is the threshold frequency for potassium?
Answer in units of Hz.
Answer in units of eV.
What is the work function of potassium? The speed of light is 3 × 108 m/s and Planck’s constant is 6.63×10−34J·s.
What is the threshold frequency for potassium?
Answer in units of Hz.
Answer in units of eV.
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c = f.λ ---------------> f = c / λ
E = hf - φ -------------> sub in 'f'
E = (hc / λ) - φ -------------------> solve for φ
φ = (hc / λ) - E
φ = [(6.63×10^−34)(3x10^8) / (350x10^-9)(1.6x10^-19)] - 1.3
[divided the (hc / λ) by 'e' to get into eV]
φ = 3.55 - 1.3 = 2.25 eV
hf = 2.25 eV
f = 2.25 eV / h = (2.25 x 1.6x10^-19) / 6.63x10^-34 = 5.4 x 10^14 Hz
I guess you know what all the symbols mean.
E = hf - φ -------------> sub in 'f'
E = (hc / λ) - φ -------------------> solve for φ
φ = (hc / λ) - E
φ = [(6.63×10^−34)(3x10^8) / (350x10^-9)(1.6x10^-19)] - 1.3
[divided the (hc / λ) by 'e' to get into eV]
φ = 3.55 - 1.3 = 2.25 eV
hf = 2.25 eV
f = 2.25 eV / h = (2.25 x 1.6x10^-19) / 6.63x10^-34 = 5.4 x 10^14 Hz
I guess you know what all the symbols mean.