Respuesta :
Answer:
932.44 km/s
Explanation:
Given that:
The work function of the magnesium = 2.3 eV
Energy in eV can be converted to energy in J as:
1 eV = 1.6022 × 10⁻¹⁹ J
So, work function = [tex]2.3\times 1.6022\times 10^{-19}\ J=3.68506\times 10^{-19}\ J[/tex]
Using the equation for photoelectric effect as:
[tex]E=\psi _0+\frac {1}{2}\times m\times v^2[/tex]
Also, [tex]E=\frac {h\times c}{\lambda}[/tex]
Applying the equation as:
[tex]\frac {h\times c}{\lambda}=\psi _0+\frac {1}{2}\times m\times v^2[/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]
m is the mass of electron having value [tex]9.11\times 10^{-31}\ kg[/tex]
[tex]\lambda[/tex] is the wavelength of the light being bombarded
[tex]\psi _0=Work\ function[/tex]
v is the velocity of electron
Given, [tex]\lambda=260\ nm=260\times 10^{-9}\ m[/tex]
Thus, applying values as:
[tex]\frac{6.626\times 10^{-34}\times 3\times 10^8}{260\times 10^{-9}}=3.68506\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2[/tex]
[tex]3.68506\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{260\times \:10^{-9}}[/tex]
[tex]\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=7.64538\times 10^{-19}-3.68506\times 10^{-19}[/tex]
[tex]\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=3.96032462\times 10^{-19}[/tex]
[tex]v^2=0.869446\times 10^{12}[/tex]
v = 9.3244 × 10⁵ m/s
Also, 1 m = 0.001 km
So, v = 932.44 km/s
