Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

If the intensity of $X-$rays becomes $\dfrac{I}{3}$ from $I$ after travelling $3.5\text{ cm}$ inside a target, then it’s intensity after travelling next $7\text{ cm}$ will be
A. $\dfrac{I}{6}$
B. $\dfrac{I}{12}$
C. $\dfrac{I}{9}$
D. $\dfrac{I}{27}$

seo-qna
Last updated date: 14th Jul 2024
Total views: 347.1k
Views today: 3.47k
Answer
VerifiedVerified
347.1k+ views
Hint: 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 $X-$rays. In this question we are not given the value of absorption coefficient of the $X-$rays, hence first we will have to find that and then proceed further.

Formula used:
$I={{I}_{0}}{{e}^{-\mu x}}$
Here $I$ is the final intensity of $X-$rays after travelling a distance $x$.
${{I}_{0}}$ is the initial intensity of the $X-$rays.
And $\mu $ is the absorption coefficient of the $X-$rays.

Complete step-by-step solution:
We know that the final intensity of $X-$rays after travelling a distance $x$ is given by the following formula:
$I={{I}_{0}}{{e}^{-\mu x}}$
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 $X-$rays becomes $\dfrac{I}{3}$ from $I$ after travelling $3.5\text{ cm}$ inside a target, hence on substituting the values in the formula we get:
$\begin{align}
  & \dfrac{I}{3}=I{{e}^{-3.5\mu }} \\
 & \Rightarrow \dfrac{1}{3}={{e}^{-3.5\mu }} \\
 & \Rightarrow {{e}^{3.5\mu }}=3 \\
 & \Rightarrow 3.5\mu =\ln 3 \\
 & \Rightarrow \mu =\dfrac{\ln 3}{3.5} \\
 & \Rightarrow \mu =0.31389\text{ c}{{\text{m}}^{-1}} \\
\end{align}$
Now that we have calculated the value of the absorption coefficient, we can the intensity of $X-$rays after travelling next $7\text{ cm}$. The total distance that the $X-$rays would travel will be:
$\begin{align}
  & x=\left( 3.5+7 \right)\text{ cm} \\
 & \Rightarrow \text{x=10}\text{.5 cm} \\
\end{align}$
On substituting the values in the formula, we get:
$\begin{align}
  & I'=I{{e}^{-0.31389\times 10.5}} \\
 & \Rightarrow I'=I{{e}^{-3.2958}} \\
 & \therefore I'=\dfrac{I}{27} \\
\end{align}$
The intensity of $X-$rays after travelling next $7\text{ cm}$will be $\dfrac{I}{27}$. Hence, the correct option is $D$.

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 $X-$rays and the absorption coefficient. The final intensity is not affected by any other factors.