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

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 =  $\frac{(3.0 * 10^{8} )(6.63 * 10^{-34} )}{700 * 10^{-9} }$

E = 2.84 x $10^{-19}$J

when

λ = 1400nm = 1400 x $10^{-9}$m

E = Ch/λ

E =  $\frac{(3.0 * 10^{8} )(6.63 * 10^{-34} )}{1400 * 10^{-9} }$

E = 1.42 x $10^{-19}$J

FOR IR-C

when

λ = 3000nm = 3000 x $10^{-9}$m

E = Ch/λ

E =  $\frac{(3.0 * 10^{8} )(6.63 * 10^{-34} )}{3000 * 10^{-9} }$

E = 6.63 x $10^{-20}$J

when

λ = 1,000,000nm = 1,000,000 x $10^{-9}$m

E = Ch/λ

E =  $\frac{(3.0 * 10^{8} )(6.63 * 10^{-34} )}{1000000 * 10^{-9} }$

E = 1.99 x $10^{-22}$J