A photosensitive metallic surface has a work function of 3.3 eV. Which of the following frequencies of incident radiation will not result in the emission of photoelectrons? (Take Planck's constant h = 6.6 imes 10⁻³⁴ J s and 1 eV = 1.6 imes 10⁻¹⁹ J)

Step-by-step solution

The work function of the metallic surface is given as = 3.3 eV. Converting the work function into Joules: = 3.3 1.6 10⁻¹⁹ J = 5.28 10⁻¹⁹ J The threshold frequency ₀ is given by the relation ₀ = h . Substituting the given values: ₀ = 5.28 10⁻¹⁹ 6.6 10⁻³⁴ Hz = 0.8 10¹⁵ Hz = 8.0 10¹⁴ Hz For photoelectric emission to take place, the frequency of the incident radiation must be greater than or equal to the threshold frequency ( ₀ ). Comparing the given options with the threshold frequency, 7.0 10¹⁴ Hz is less than 8.0 10