Question

asked Sep 22, 2020 168k views

1 Answer

Dzeikei · Answer 1 · 2020-09-27T19:52:36+0000

Answers:

The energy
of a photon is given by the following formula:

E=h.f (1)

Where:

h=4.136(10)^(-15) eV.s is the Planck constant

is the frequency in hertz
Hz=s^(-1)

Now, the frequency has an inverse relation with the wavelength
$\lambda$ :

$f=(c)/(\lambda)$ (2)

Where
c=3(10)^(8)m/s is the speed of light in vacuum

Substituting (2) in (1):

$E=(hc)/(\lambda)$ (3)

Knowing this, let's begin with the answers:

For
$\lambda=437nm=437(10)^(-9)m$

E=((4.136(10)^(-15) eV.s)(3(10)^(8)m/s))/(437(10)^(-9)m)

E=(1.24(10)^(-6)eV.m )/(437(10)^(-9)m)

E=2.837eV

For
$\lambda=533nm=533(10)^(-9)m$

E=(1.24(10)^(-6)eV.m )/(533(10)^(-9)m)

E=2.327eV

For
$\lambda=564nm=564(10)^(-9)m$

E=(1.24(10)^(-6)eV.m )/(564(10)^(-9)m)

E=2.2eV