The energy [tex]E[/tex] of a photon is given by the following formula:
[tex]E=h.f[/tex] (1)
Where:
[tex]h=4.136(10)^{-15} eV.s[/tex] is the Planck constant
[tex]f[/tex] is the frequency in hertz [tex]Hz=s^{-1}[/tex]
Now, the frequency has an inverse relation with the wavelength [tex]\lambda[/tex]:
[tex]f=\frac{c}{\lambda}[/tex] (2)
Where [tex]c=3(10)^{8}m/s[/tex] is the speed of light in vacuum
Substituting (2) in (1):
[tex]E=\frac{hc}{\lambda}[/tex] (3)
Knowing this, let's begin with the answers:
For [tex]\lambda=437nm=437(10)^{-9}m[/tex]
[tex]E=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{437(10)^{-9}m}[/tex]
[tex]E=\frac{1.24(10)^{-6}eV.m }{437(10)^{-9}m}[/tex]
[tex]E=2.837eV[/tex]
For [tex]\lambda=533nm=533(10)^{-9}m[/tex]
[tex]E=\frac{1.24(10)^{-6}eV.m }{533(10)^{-9}m}[/tex]
[tex]E=2.327eV[/tex]
For [tex]\lambda=564nm=564(10)^{-9}m[/tex]
[tex]E=\frac{1.24(10)^{-6}eV.m }{564(10)^{-9}m}[/tex]
[tex]E=2.2eV[/tex]
