Respuesta :
Answer:
275.3 nm is the wavelength of light required for mercury.
Mercury can not be used to generate electricity from the sun because wavelength at which mercury will emit an electron is smaller than 500 nm.
Explanation:
The wavelength of light required for mercury to emit an electron.
The wavelength of the radiation = [tex]\lambda [/tex]
Energy required fro mercury to to emit an electron = E
Energy required fro mercury to to emit an electron will the energy if the radiation = E' = [tex]7.22\times 10^{-19} J[/tex]
E' = E
To calculate the wavelength of light, we use the equation:
[tex]E=\frac{hc}{\lambda }[/tex]
where,
[tex]\lambda[/tex] = wavelength of the light
h = Planck's constant = [tex]6.626\times 10^{-34} Js[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda =\frac{hc}{E}[/tex]
[tex]=\frac{6.626\times 10^{-34} Js\times 3\times 10^8m/s}{7.22\times 10^{-19} J}[/tex]
[tex]\lambda =2.753\times 10^{-7} m=2.753\times 10^{-7}\times 10^ nm =275.3 nm[/tex]
Wavelength of the sun light in the visible region = 500 nm
500 nm > 275.3 nm
[tex]E\propto \frac{1}{\lambda }[/tex]
Less energy < more energy
So, this means that mercury can not be used to generate electricity from the sun.
The wavelength of light in nm is
- 275nm
The mathematical formula for wavelength
[tex]\lambda = \frac{hc}{E}\\\\ \lambda = \frac{6.626*10^{-34} * 3*10^8}{7.22*10^{-19}}\\\\ \lambda = 2.7531856*10^{−7}\\\\ \lambda = 275nm [/tex]
No, the therhold energy is larger than the wavelength of sun, therefore, electricity will not be generated.
For more information on wavelength, visit
https://brainly.com/question/10728818
