Question

The percentage of transmittance of a solution of X at 300 nm and 25 C is 20% for a 5x10^-4 M solution in a 1-cm cell. a) Calculate the absorbance A and the molar absorption coefficient,ε .
b) Calculate the energy in joules per mole, and electron volt (eV) of photons of wavelength 300 nm.
(h=6.62x10^-34 J s), (c=3x10^8 m s^-1), (1 eV=1.6x10^-19 J), (1 nm =10^-9 m)

Answer 1

The absorbance of the solution is 0.7. The  molar absorption coefficient is  1400 M-1cm-1. The energy of the photon is  4.1 eV.

To find the absorbance;

A = log10(100) - log 10(%T)

A = 2 - log10(20)

A = 0.7

Since;

A = εcl

ε = molar absorption coefficient

c = concentration

l= path length

ε = A/cl

ε = 0.7/ 5 x 10^-4 M x 1 cm

ε = 1400 M-1cm-1

b) From

E = hc/λ

E = energy

h = Plank's constant

c = speed of light

λ = wavelength

E = 6.6 × 10^-34 Js × 3 × 10^8/300 ×  10^-9

E = 6.6 ×  10^-19 J

However;

1 eV = 1.6x10^-19 J

x eV =  6.6 ×  10^-19 J

x = 6.6 ×  10^-19 J × 1 eV/ 1.6x10^-19 J

x = 4.1 eV

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