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Question: If the intensity of \(X-\)rays becomes \(\dfrac{I}{3}\) from \(I\) after travelling \(3.5\text{ cm}\...

If the intensity of XX-rays becomes I3\dfrac{I}{3} from II after travelling 3.5 cm3.5\text{ cm} inside a target, then it’s intensity after travelling next 7 cm7\text{ cm} will be
A. I6\dfrac{I}{6}
B. I12\dfrac{I}{12}
C. I9\dfrac{I}{9}
D. I27\dfrac{I}{27}

Explanation

Solution

To solve these types of questions we need to know about the relationship between initial intensity, final intensity, absorption coefficient and the distance travelled by XX-rays. In this question we are not given the value of absorption coefficient of the XX-rays, hence first we will have to find that and then proceed further.

Formula used:
I=I0eμxI={{I}_{0}}{{e}^{-\mu x}}
Here II is the final intensity of XX-rays after travelling a distance xx.
I0{{I}_{0}} is the initial intensity of the XX-rays.
And μ\mu is the absorption coefficient of the XX-rays.

Complete step-by-step solution:
We know that the final intensity of XX-rays after travelling a distance xx is given by the following formula:
I=I0eμxI={{I}_{0}}{{e}^{-\mu x}}
In the question, the value of absorption coefficient is not given. Hence our first task would be to calculate the value of absorption coefficient. We know that the intensity of XX-rays becomes I3\dfrac{I}{3} from II after travelling 3.5 cm3.5\text{ cm} inside a target, hence on substituting the values in the formula we get:
I3=Ie3.5μ 13=e3.5μ e3.5μ=3 3.5μ=ln3 μ=ln33.5 μ=0.31389 cm1 \begin{aligned} & \dfrac{I}{3}=I{{e}^{-3.5\mu }} \\\ & \Rightarrow \dfrac{1}{3}={{e}^{-3.5\mu }} \\\ & \Rightarrow {{e}^{3.5\mu }}=3 \\\ & \Rightarrow 3.5\mu =\ln 3 \\\ & \Rightarrow \mu =\dfrac{\ln 3}{3.5} \\\ & \Rightarrow \mu =0.31389\text{ c}{{\text{m}}^{-1}} \\\ \end{aligned}
Now that we have calculated the value of the absorption coefficient, we can the intensity of XX-rays after travelling next 7 cm7\text{ cm}. The total distance that the XX-rays would travel will be:
x=(3.5+7) cm x=10.5 cm \begin{aligned} & x=\left( 3.5+7 \right)\text{ cm} \\\ & \Rightarrow \text{x=10}\text{.5 cm} \\\ \end{aligned}
On substituting the values in the formula, we get:
I=Ie0.31389×10.5 I=Ie3.2958 I=I27 \begin{aligned} & I'=I{{e}^{-0.31389\times 10.5}} \\\ & \Rightarrow I'=I{{e}^{-3.2958}} \\\ & \therefore I'=\dfrac{I}{27} \\\ \end{aligned}
The intensity of XX-rays after travelling next 7 cm7\text{ cm}will be I27\dfrac{I}{27}. Hence, the correct option is DD.

Note: To solve these types of questions we need to remember the relation between final intensity and initial intensity that depends on various factors like distance travelled by XX-rays and the absorption coefficient. The final intensity is not affected by any other factors.