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Many calculators use photocells as their energy source. Find the maximum wavelength needed to remove an electron from silver (∅ = 7.59×10⁻¹⁹J ). Is silver a good choice for a photocell that uses visible light? [The concept of the work function (∅ ) is explained in Problem 7.69.]

Respuesta :

  • The maximum wavelength needed to remove an electron from silver is 261 nm.
  • And silver is not a good choice for a photocell that uses visible light.

Calculation of wavelength:

E = h x c/ λ

Where,  

h = Plank's constant 6.626 x [tex]10^{-34}[/tex] Js

c = speed of light 3 x [tex]10^{8}[/tex] m/s

λ = wavelength of the light

Energy = 7.59×10⁻¹⁹J

7.59×10⁻¹⁹J =  (6.626 x [tex]10^{-34}[/tex]) x (3 x [tex]10^{8}[/tex]) / λ

λ = 1.99 x [tex]10^{20}[/tex]/ 7.59×10⁻¹⁹

λ = 2.61 x [tex]10^{-9}[/tex]

= 261 nm

Hence, the maximum wavelength needed to remove an electron from silver is 261 nm.

Spectrum for the visible range extends from 380 to 740 nm.

Silver is not a wise choice, therefore.

Learn more about wavelength here:

https://brainly.com/question/13533093

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