Answer:
The energy of an individual photon that comes out of the laser pointer is 1.91 eV
Explanation:
The energy of a photon can be obtained using the expression below
E = hc/λ
where E is the energy;
h is the Planck's constant = 6.626 x 10-34 Js;
c is the speed of light = 3.00 x [tex]10^{8}[/tex] m/s (speed of light);
λ is the wavelength = 650 nm =650 x [tex]10^{-9}[/tex] m.
E = (6.626 x 10-34 Js) x (3.00 x [tex]10^{8}[/tex] m/s) /650 x [tex]10^{-9}[/tex] m
E = 3.058 x [tex]10^{-19}[/tex] J
1 joule = 6.242 x [tex]10^{18}[/tex] eV
3.058 x [tex]10^{-19}[/tex] J = 3.058 x [tex]10^{-19}[/tex] J x 6.242 x [tex]10^{18}[/tex] eV = 1.91 eV
Therefore the energy of an individual photon that comes out of the laser pointer is 1.91 eV
