Question

Calculate the energy of a photon having a wavelength in thefollowing ranges.
(a) microwave, with λ = 50.00 cm
eV
(b) visible, with λ = 500 nm
eV
(c) x-ray, with λ = 0.50 nm
eV

Answer 1

Answers:

The energy $E$ 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

$f$ is the frequency

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

$f=\frac{c}{\lambda}$ (2)

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

Substituting (2) in (1):

$E=\frac{hc}{\lambda}$ (3)

Knowing this, let's begin with the answers:

(a) Microwave: 50.00 cm

For $\lambda=50cm=0.5m$

$E=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{0.5m}$  

$E=\frac{1.24(10)^{-6}eV.m }{0.5m}$  

$E=2.48(10)^{-6}eV$  

(b) Visible: 500 nm

For $\lambda=500nm=500(10)^{-9}m$

$E=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{500(10)^{-9}m}$  

$E=\frac{1.24(10)^{-6}eV.m }{500(10)^{-9}m}$  

$E=2.48 eV$  

(c) X-ray: 0.5 nm

For $\lambda=0.5nm=0.5(10)^{-9}m$

$E=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{0.5(10)^{-9}m}$  

$E=\frac{1.24(10)^{-6}eV.m }{0.5(10)^{-9}m}$  

$E=2480 eV$  

As we can see, as the wavelength decreases, the energy increases.