Question

A certain photochemical reaction requires an excitation energy of 126 kJ mol^-1 . To what values does this correspond in the following units: (A) frequency of light, (B) wave number, (C) wavelength in nanometers, and (D) electron volts?

Answer #1

Given that excitation energy = 126 kJ mol^{-1}

^{ } E = 126 kJ mol^{-1} = 126000
J/mol

Divide this energy with Avogadro number to get energy in J.

E = 126000 J mol^{-1} / Avogadro number = 126000 J
mol^{-1} / 6.023 x 10^{23} mol^{-1}

^{ } = 20919.8 x 10^{-23} J

E = 20919.8 x 10^{-23} J

A) Frequency:

E = hv , v = frequency

v = E/h

= 20919.8 x 10^{-23} J / 6.626 x
10^{-34} J.s

= 3157.2 x 10^{11} Hz

Therefore, frequency v = 3157.2 x 10^{11}
Hz

C)

E = hc /**λ , λ =** wavelength

λ = hc/E

** =** 6.626 x 10^{-34} J.s x 3
x10^{8} m/s / 20919.8 x 10^{-23} J

= 950 x 10^{-9} m

= 950 nm

Therefore, wavelength = 950 nm

B)

wavenumber = 1/ wavelength

= 1/ 950 x 10^{-9} m

= 0.00105 x 10^{9} m^{-1}

Therefore, wavenumber = 0.00105 x 10^{9}
m^{-1}

D) 1 J = 6.24 x 10^{18} ev

E = 20919.8 x 10^{-23} J

= 20919.8 x 10^{-23} x 6.24 x 10^{18}

^{ } = 130539 x 10^{-5} eV

E = 130539 x 10^{-5} eV

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