Assume that the exponent fix in the equation I I0emux is eq

Assume that the exponent fix in the equation I = I_0e^-mux is equal to or less than 0.1. If we consider an error less than 1%, the number of photons transmitted can be derived as I = I_0(1 - mux) and the number of photons attenuated I_at = I_0mux Electrons with kinetic energy of 1.0 MeV have a specific ionization in air of about 6000 IP/m. What is the LET of these electrons in air? It is hypothesized that the number of text messages send by an individual is related to the size of the brain tumor developed in that individual. However, we know very well that these two events are unrelated. Draw the ROC curve for this particular situation

Solution

e ^{– x} = 1- x+ (x)²/2- ….

As x is less than 0.1 therefore higher terms of series can be neglected. Thus,

e ^{– x} = 1- x

Substitute this value in equation of I(x) = I_O e ^{– x}

I(x) = I_O (1- x)

Thus, number of photons transmitted is I(t) = I_O (1- x).

Number of photons attenuated is I_O- I(t) that is total number of photons incident – number of photons transmitted.

I_O - I_O (1- x) = I_O x.

2) •The linear energy transfer (LET) is the average loss in energy per unit length of path of the incident radiation

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•The LET is the product of the specific ionization and the W-quantity

LET = (SI)(W)

W –quantity or W, is an average energy of 33.85 eV expended by charged particles per ion pair produced in air

Therefore LET for electrons can be given as

         

LET = (6000 IP/m)(33.85 ev/IP)

LET = 203100 eV