Respuesta :
Answer:
[tex]\lambda=540.16\ nm[/tex]
Explanation:
Given that:
The energy of the photon = 2.3 eV
Energy in eV can be converted to energy in J as:
1 eV = 1.60 × 10⁻¹⁹ J
So, Energy = [tex]2.3\times 1.60\times 10^{-19}\ J=3.68\times 10^{-19}\ J[/tex]
Considering
[tex]Energy=\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 being bombarded
Thus,
[tex]3.68\times 10^{-19}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]
[tex]\frac{3.68}{10^{19}}=\frac{19.878}{10^{26}\lambda}[/tex]
[tex]3.68\times \:10^{26}\lambda=1.9878\times 10^{20}[/tex]
[tex]\lambda=5.40163\times 10^{-7}\ m=540.16\times 10^{-9}\ m[/tex]
Also,
1 m = 10⁻⁹ nm
So,
[tex]\lambda=540.16\ nm[/tex]
