Question

Two of the types of infrared light, IR-C and IR-A, are both components of sunlight. Their wavelengths range from 3000 to 1,000,000 nmnm for IR-C and from 700 to 1400 nmnm for IR-A. Compare the energy of microwaves, IR-C, and IR-A.

Answer 1

Answer: IR-A possess more energy than IR-C

Step-by-step explanation:

Before anything, you need to understand that; the longer the wavelength a wave possess, the smaller the energy it has.

We have to use planck's equation and C=λv

where; C = Speed of light (3.0 x
`10^(8)`m/s)

λ = wave length

v = frequency

planck's equation = E = hv

Equating the two equations; we get

E/h = C/λ

E = Ch/λ

where;

h = planck's constant (6.63 x
`10^(-34)`Js)

Comparing the energies of microwaves,

FOR IR-A

when

λ = 700nm = 700 x
`10^(-9)`m

E = Ch/λ

E =
`((3.0 * 10^(8) )(6.63 * 10^(-34) ))/(700 * 10^(-9) )`

E = 2.84 x
`10^(-19)`J

when

λ = 1400nm = 1400 x
`10^(-9)`m

E = Ch/λ

E =
`((3.0 * 10^(8) )(6.63 * 10^(-34) ))/(1400 * 10^(-9) )`

E = 1.42 x
`10^(-19)`J

FOR IR-C

when

λ = 3000nm = 3000 x
`10^(-9)`m

E = Ch/λ

E =
`((3.0 * 10^(8) )(6.63 * 10^(-34) ))/(3000 * 10^(-9) )`

E = 6.63 x
`10^(-20)`J

when

λ = 1,000,000nm = 1,000,000 x
`10^(-9)`m

E = Ch/λ

E =
`((3.0 * 10^(8) )(6.63 * 10^(-34) ))/(1000000 * 10^(-9) )`

E = 1.99 x
`10^(-22)`J