Question

What is the energy, in kJ/mole, associated with photons having the following wavelength

A. 280nm

B. 400nm

C. 750nm

D. 4000nm

What is the significane of each of these wavelengths

Answer 1

A)

lambda = 180 nm = 2.80*10^-7 m

Find energy of 1 photon first

use:

E = h*c/lambda

=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(2.80*10^-7 m)

= 7.099*10^-19 J

This is energy of 1 photon

Energy of 1 mol = energy of 1 photon * Avogadro's number

= 7.099*10^-19*6.022*10^23 J/mol

= 4.275*10^5 J/mol

= 4.275*10^2 KJ/mol

Answer: 4.275*10^2 KJ/mol

b)

lambda = 400 nm = 4.00*10^-7 m

Find energy of 1 photon first

use:

E = h*c/lambda

=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(4.00*10^-7 m)

= 4.969*10^-19 J

This is energy of 1 photon

Energy of 1 mol = energy of 1 photon * Avogadro's number

= 4.969*10^-19*6.022*10^23 J/mol

= 2.993*10^5 J/mol

= 2.993*10^2 KJ/mol

Answer: 2.993*10^2 KJ/mol

c)

lambda = 750 nm = 7.50*10^-7 m

Find energy of 1 photon first

use:

E = h*c/lambda

=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(7.50*10^-7 m)

= 2.65*10^-19 J

This is energy of 1 photon

Energy of 1 mol = energy of 1 photon * Avogadro's number

= 2.65*10^-19*6.022*10^23 J/mol

= 1.596*10^5 J/mol

= 1.596*10^2 KJ/mol

Answer: 1.596*10^2 KJ/mol

D)

lambda = 4000 nm = 4.000*10^-6 m

Find energy of 1 photon first

use:

E = h*c/lambda

=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(4.000*10^-6 m)

= 4.97*10^-20 J

This is energy of 1 photon

Energy of 1 mol = energy of 1 photon * Avogadro's number

= 4.97*10^-20*6.022*10^23 J/mol

= 2.993*10^4 J/mol

= 29.93 KJ/mol

Answer: 29.93 KJ/mol

significance: 280 nm is in ultravilet region, 4000 nm is in infrared region. Other 2 are in visible region