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The work function of a certain metal is φ = 3.55 eV. Determine the minimum frequency of light f0 for which photoelectrons are emitted from the metal. (Planck's constant is: h = 4.1357×10-15 eVs.)

Respuesta :

Answer:

Explanation:

Let f₀ be the frequency .

energy of photons having frequency of f₀

= hf₀ where h is plank's constant

for electron to get ejected , work function should be equal to energy of photon

hf₀ = 3.55

4.1357 x 10⁻¹⁵ x f₀ = 3.55

f₀ = 8.58 x 10¹⁴ Hz .

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