Answered

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (with an incident intensity, lo) after it has traveled through 1 cm of the material? After it has traveled through 2 cm of the material? (b) What is the optical density of a 1 cm thick piece of such a material? Of a 2 cm thick piece of such a material?

Answer :

fenixbm

Answer:

a.

  • After it has traveled through 1 cm : [tex]I(1 \ cm) = 0.5488 I_0 [/tex]
  • After it has traveled through 2 cm : [tex]I(2 \ cm) = 0.3012 I_0 [/tex]

b.

  • After it has traveled through 1 cm : [tex]od( 1\ cm) =  0.2606[/tex]
  • After it has traveled through 2 cm :  [tex]od( 2\ cm) =  0.5211[/tex]

Explanation:

a.

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient [tex]\mu[/tex] the formula is:

[tex]I(x) = I_0 e^{-\mu x}[/tex]

where I is the intensity of the beam, [tex]I_0[/tex] is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

[tex]I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}[/tex]

[tex]I(1 \ cm) = I_0 e^{- 0.6}[/tex]

[tex]I(1 \ cm) = I_0 e^{- 0.6}[/tex]

[tex]I(1 \ cm) = 0.5488 \ I_0 [/tex]

After travelling 2 cm:

[tex]I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}[/tex]

[tex]I(2 \ cm) = I_0 e^{- 1.2}[/tex]

[tex]I(2 \ cm) = I_0 e^{- 1.2}[/tex]

[tex]I(2 \ cm) = 0.3012 \ I_0 [/tex]

b

The optical density od is given by:

[tex]od(x) = - log_{10} ( \frac{I(x)}{I_0} )[/tex].

So, after travelling 1 cm:

[tex]od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )[/tex]

[tex]od( 1\ cm) = - log_{10} ( 0.5488 )[/tex]

[tex]od( 1\ cm) = - (  - 0.2606)[/tex]

[tex]od( 1\ cm) =  0.2606[/tex]

After travelling 2 cm:

[tex]od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )[/tex]

[tex]od( 2\ cm) = - log_{10} ( 0.3012 )[/tex]

[tex]od( 2\ cm) = - (  - 0.5211)[/tex]

[tex]od( 2\ cm) =  0.5211[/tex]

Other Questions