Question

Two of the types of infrared light, IR-B and IR-A, are both components of sunlight. Their wavelengths range from 1400 to 3000 nm for IR-B and from 700 to 1400 nm for IR-A. Compare the energy of microwaves, IR-B, and IR-A. Rank from greatest to least energy per photon. To rank items as equivalent, overlap them.

Answer 1

Answer:

The IR-A is the greatest energy per photon compare to that of IR-B.

Explanation:

Given that,

Wavelength of (IR-B)= 1400 nm

Wavelength of (IR-A)= 700 nm

We need to calculate the energy for IR-B

Using formula of energy

$E=\dfrac{hc}{\lambda}$

Put the value into the formula

$E=\dfrac{6.62\times10^{-34}\times3\times10^{8}}{1400\times10^{-9}}$

$E=1.418\times10^{-19}\ J$

The energy for IR-B is $1.418\times10^{-19}\ J$

We need to calculate the energy for IR-A

Using formula of energy

$E=\dfrac{hc}{\lambda}$

Put the value into the formula

$E=\dfrac{6.62\times10^{-34}\times3\times10^{8}}{700\times10^{-9}}$

$E=2.83\times10^{-19}\ J$

The energy for IR-B is $2.83\times10^{-19}\ J$

Hence, The IR-A is the greatest energy per photon compare to that of IR-B.