Question

A liquid is exposed to infrared radiation with a wavelength of 6.92 × 10 − 4 cm. 6.92×10−4 cm. Assume that all the radiation is absorbed and converted to heat. How many photons are required for the liquid to absorb 58.28 J 58.28 J of heat?

Answer 1

Answer:

The no. of photons required for total heat absorbed of 58.28 J is N = 20.28  × $10^{22}$ Photons

Explanation:

Wave length $\lambda$ = 6.92 × $10^{-4}$ cm =  6.92 × $10^{-6}$ m

Total Energy E = 58.28 J

Energy content of a single photon is given by

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

where h = plank constant = 6.626 × $10^{-34}$ J - s

c = speed of the light = 3 × $10^{8}$ $\frac{m}{s}$

So Energy content of a single photon is

$E = \frac{(6.626) (3)}{6.92}$ × $10^{-22}$

E = 2.87 × $10^{-22}$ J

No. of photons required is

$N = \frac{Total\ energy}{energy \ of \ a \ single \ photon}$

N = $\frac{58.28}{2.87}$ × $10^{22}$

N = 20.28  × $10^{22}$ Photons

This is the no. of photons required for total heat absorbed of 58.28 J