Many calculators use photocells to provide their energy. Find the maximum wavelength needed to remove an electron from silver (Φ = 7.59 x 10⁻¹⁹ J). Is silver a good choice for a photocell that uses visible light?

Respuesta :

Answer:

[tex]\lambda=2.61\times 10^{-9}\ m[/tex] = 261 nm

Silver is not a good choice.

Explanation:

[tex]E=\frac {h\times c}{\lambda}[/tex]

Where,  

h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]

c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]

[tex]\lambda[/tex] is the wavelength of the light

Given that:- Energy = [tex]7.59\times 10^{-19}\ J[/tex]

[tex]7.59\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]

[tex]7.59\times \:10^{26}\times \lambda=1.99\times 10^{20}[/tex]

[tex]\lambda=2.61\times 10^{-9}\ m[/tex] = 261 nm

Visible range has a spectrum of 380 to 740 nm

So, Silver is not a good choice.