Question

Determine the energy of 2.00 mol of photons for each of the following kinds of light.

Part A:   infrared radiation (1410 nm ) E= ___kJ

Part B:   visible light (485 nm ) E= ___kJ

Part C:   ultraviolet radiation (170 nm ) E= ___kJ

Answer 1

a)

the formula

E = hc/WL

h = 6.626*10^-34

c = 3*10^8

WL = 1410 nm = 1410*10^-9

then

E = (6.626*10^-34)( 3*10^8)/(1410*10^-9) = 1.4097872*10^-19 J/photon

we have 2 mol so

E = (1.4097872*10^-19)(2*6.022*10^23) = 169794.770368 J = 169.795 kJ

B)

the formula

E = hc/WL

h = 6.626*10^-34

c = 3*10^8

WL = 1410 nm = 1410*10^-9

then

E = (6.626*10^-34)( 3*10^8)/(485*10^-9) = 4.0985567*10^-19 J/photon

we have 2 mol so

E = (4.0985567*10^-19)(2*6.022*10^23) = 493630.168948 J = 493.63 kJ

c)

the formula

E = hc/WL

h = 6.626*10^-34

c = 3*10^8

WL = 1410 nm = 1410*10^-9

then

E = (6.626*10^-34)( 3*10^8)/(170 *10^-9) = 1.1692*10^-19 J/photon

we have 2 mol so

E = (1.1692*10^-19)(2*6.022*10^23) = 140818.448 J = 140.818.4 kJ