The intensity of gamma radiation emitted from | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The intensity of radiation after passing through a material of thickness $x$ is given by the formula: $I = I_0 e^{-kx}$,where $k$ is the absorptio

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The intensity of radiation after passing through a material of thickness $x$ is given by the formula: $I = I_0 e^{-kx}$,where $k$ is the absorption coefficient.From the given data,when $x_1 = 36 \, mm$,$I_1 = I_0/8$.Substituting these values: $I_0/8 = I_0 e^{-k(36)} \Rightarrow 1/8 = e^{-36k}$.Taking the natural logarithm on both sides: $\ln(1/8) = -36k \Rightarrow -\ln(8) = -36k \Rightarrow \ln(2^3) = 36k \Rightarrow 3 \ln(2) = 36k \Rightarrow k = \frac{3 \ln(2)}{36} = \frac{\ln(2)}{12}$.Now,we need to find the thickness $x_2$ when $I_2 = I_0/2$.Using the formula: $I_0/2 = I_0 e^{-kx_2} \Rightarrow 1/2 = e^{-kx_2}$.Taking the natural logarithm: $\ln(1/2) = -kx_2 \Rightarrow -\ln(2) = -kx_2 \Rightarrow \ln(2) = kx_2$.Substituting the value of $k$: $\ln(2) = (\frac{\ln(2)}{12}) x_2$.Solving for $x_2$: $x_2 = 12 \, mm$.

The intensity of gamma radiation emitted from a source is $I_0$. After passing through a lead sheet of thickness $36 \, mm$,its intensity becomes $I_0/8$. Find the thickness of the lead sheet in $mm$ when the intensity becomes $I_0/2$.

Similar Questions

The nucleus $_{48}^{115}Cd$ after two successive ${\beta ^ - }$ decays will give:

${ }_{82}^{290} X \xrightarrow{\alpha} Y \xrightarrow{e^{+}} Z \xrightarrow{\beta^{-}} P \xrightarrow{e^{-}} Q$
In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are

Obtain the maximum kinetic energy of $\beta$-particles and the radiation frequencies of $\gamma$-decays in the decay scheme shown in the figure. You are given that:
$m(^{198}Au) = 197.968233 \; u$
$m(^{198}Hg) = 197.966760 \; u$

$A$ radioactive element $A$ decays into radioactive element $C$ by the following processes in succession.
$A \rightarrow B + {}_{2}^{4}He$
$B \rightarrow C + 2e^{-}$
Then elements

The energy spectrum of $\beta$-particles [number $N(E)$ as a function of $\beta$-energy $E$] emitted from a radioactive source is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module