Question

A sample in a 1-cm cuvette is assessed with a spectrophotometer set to 620 nm. If the sample returns a transmittance of 62%, then what is the approximate absorbance of the sample?.

Answer 1

The absorbance of the given sample is 0.2076.

Length of the cuvette = 1 cm

Wavelength of the spectrophotometer = 620nm

Transmittance of the sample = 62%

We know, absorbance is related to transmittance by the formula written below;

A = log(1/T)                                                 equation 1

Also, percent transmittance %T = 100T

or we can write, T = (%T) / 100

On substituting value of T in equation 1 , we get

A = log(100/(%T))    

⇒ A = log(100) - log (%T)      

⇒ A = log($10^{2}$) - log(%T)

⇒ A=  2log(10) - log(%T)    

We know, log10 = 1. On substituting the value of log 10 in above equation, we get

A = 2 - log(%T)

⇒ A = 2 - log(62)

⇒ A = 2 - 1.7924

⇒ A = 0.2076

The absorbance of the given sample is 0.2076.

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