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The correct statement is
Ultraviolet light has both a higher frequency and a higher radiant energy than visible light.
because ultraviolet light has wavelength smaller than the visible light hence has a greater frequency as compared to visible light. (frequency is inversely related to wavelength. hence smaller the wavelength , greater will be the frequency)
we also know that the radiant energy is directly proportional to the frequency. hence greater the frequency , greater will be the radiant energy.
Since the frequency is greater for ultraviolet light , it radiant energy is also greater
Answer:
Ф = 2.179 eV
Explanation:
This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.
E = K + Ф
the energy of the photons is given by the Planck relation
E = h f
we substitute
h f = K + Ф
Ф= hf - K
the speed of light is related to wavelength and frequency
c = λ f
f = c /λ
Φ = 
let's reduce the energy to the SI system
K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J
calculate
Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹ -1.424 10⁻¹⁹
Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹
Ф = 3.4571 10⁻¹⁹ J
we reduce to eV
Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
Ф = 2.179 eV
This is related to the energy carried by photons of light the energy of each photon is proportional to the frequency of the light since red light has a lower frequency then violet light and photons of red light carry less energy than the photons of violet light as a result the red protons eject electrons that have less energy than the ejected electrons by Violet photons
Run electrity through or is postive to the circuit
