Question

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

Answer 1

(a)
`2.5\cdot 10^(-6)eV`

The energy of a photon is given by:

`E=(hc)/(\lambda)`

where

`h=6.63\cdot 10^(-34)Js` is the Planck constant

`c=3\cdot 10^8 m/s` is the speed of light

`\lambda` is the wavelength

For the microwave photon,

`\lambda=50.00 cm = 0.50 m`

So the energy is

`E=((6.63\cdot 10^(-34)Js)(3\cdot 10^8 m/s))/(0.50 m)=4.0\cdot 10^(-25) J`

And converting into electronvolts,

`E=(4.0\cdot 10^(-25)J)/(1.6\cdot 10^(-19) J/eV)=2.5\cdot 10^(-6)eV`

(b)
`2.5 eV`

For the energy of the photon, we can use the same formula:

`E=(hc)/(\lambda)`

For the visible light photon,

`\lambda=500 nm = 5 \cdot 10^(-7)m`

So the energy is

`E=((6.63\cdot 10^(-34)Js)(3\cdot 10^8 m/s))/(5\cdot 10^(-7) m)=4.0\cdot 10^(-19) J`

And converting into electronvolts,

`E=(4.0\cdot 10^(-19)J)/(1.6\cdot 10^(-19) J/eV)=2.5 eV`

(c)
`2500 eV`

For the energy of the photon, we can use the same formula:

`E=(hc)/(\lambda)`

For the x-ray photon,

`\lambda=0.5 nm = 5 \cdot 10^(-10)m`

So the energy is

`E=((6.63\cdot 10^(-34)Js)(3\cdot 10^8 m/s))/(5\cdot 10^(-10) m)=4.0\cdot 10^(-16) J`

And converting into electronvolts,

`E=(4.0\cdot 10^(-16)J)/(1.6\cdot 10^(-19) J/eV)=2500 eV`