Question

a certain substance in a cell of length l absorbs 10%of an incidents light. What fraction of the incidents light will be absorbed in a cell of five times as long​

Answer 1

Answer:

To convert a value from absorbance to percent transmittance, use the following equation: %T = antilog (2 – absorbance) Example: convert an absorbance of 0.505 to %T: antilog (2 – 0.505) = 31.3 %T.

Explanation:

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Answer 2

Answer:

The fraction of the incident light that is absorbed when the cell length is increased by five times is 40.98%.

Explanation:

Given the length of a cell,  $l~ units$

The incident light that is absorbed, $I_{abs}= 10\%$

Then the transmitted light, $I_t= (100 - 10)\% = 90\%$

Let the intensity of input light be $I_0$

Then the intensity of transmitted light is 90% of input intensity.

$I_t=90\%\times I_0\implies\dfrac{ I_t}{I_0} =0.90$

According to Beer-Lambert's law, the fraction of incident light absorbed by a solution at a given wavelength is given by the relation

$log\Big(\frac{I_0}{I_t} \Big)=\epsilon lc$

where $l$ is the length of the cell, $\epsilon$ is the molar extinction coefficient, c is the concentration of the cell, and  $I_0~\&~I_t$ are the incident and transmitted light intensities.

Using the formula for a cell of length $l$,

$\implies log\Bigg(\dfrac{ I_t}{I_0} \Bigg) =-\epsilon lc$

$\implies log(0.90) =-\epsilon lc$

$\implies -0.0458=-\epsilon lc$

$\implies\epsilon= \frac{0.0458}{lc}$

Given that the length is increased 5 times, thus using the formula for a cell of length $5l$,

$log\Bigg(\dfrac{ I_t}{I_0} \Bigg) =-\epsilon (5l)c$

$=-\frac{0.0458}{lc}(5l)c$

$=-0.0458\times 5=-0.229$

Taking log to the right hand side,

$\dfrac{ I_t}{I_0} = antilog(-0.0229)=0.5902$

$\implies I_t} = 0.5902\times {I_0}$

Intensity of light absorbed when cell length is increased is $I_{abs}=I_0 - 0.5902~ {I_0}$

$=(1- 0.5902)I_0 =0.4098I_0$

$=40.98\% ~ \mathbi{of} ~I_0$

Therefore, 40.98% of incident light will be absorbed if the cell length is increased by five times.