The rate of disintegration of a radioactive e | vedclass

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

Question

(D) Let $m$ be the mass of the radioactive element at time $t$. According to the problem,the rate of disintegration is proportional to its mass: $\fra

Vedclass · Accepted Answer

$\log 2$

Vedclass · Answer

(D) Let $m$ be the mass of the radioactive element at time $t$. According to the problem,the rate of disintegration is proportional to its mass: $\frac{dm}{dt} = -km$ (where $k > 0$ is the decay constant). Separating the variables,we get: $\frac{dm}{m} = -k \, dt$. Integrating both sides: $\int \frac{dm}{m} = -\int k \, dt \implies \ln(m) = -kt + C$. At $t = 0$,the initial mass $m = 6 	ext{ gm}$,so $\ln(6) = C$. Thus,$\ln(m) = -kt + \ln(6) \implies \ln(\frac{m}{6}) = -kt$. We want to find the time $t$ when the mass $m = 3 	ext{ gm}$: $\ln(\frac{3}{6}) = -kt \implies \ln(\frac{1}{2}) = -kt \implies -\ln(2) = -kt \implies t = \frac{\ln(2)}{k}$. Therefore,the time $t$ is proportional to $\log 2$.

The rate of disintegration of a radioactive element at time $t$ is proportional to its mass at that time. Then the time during which the original mass of $6 \text{ gm}$ will disintegrate into its mass of $3 \text{ gm}$ is proportional to

Similar Questions

If the slope of the tangent on a curve at any point $(x, y)$ is equal to $\frac{y^2-x^2}{2xy}$,then the equation of the normal at the point $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ is

$A$ population $P$ grows at the rate given by the equation $\frac{dP}{dt} = 0.05 P$. In how many years will the population double?

$A$ radioactive substance has a half-life of $h \ days$. Its initial decay rate is given by (Note that at $t = 0, M = m_0$):

In a bank,the principal increases continuously at the rate of $5 \%$ per year. An amount of $Rs. 1000$ is deposited in this bank. How much will it be worth after $10$ years? (Given: $e^{0.5} = 1.648$)

Given $\frac{d^2 y}{d x^2}+\cot x \frac{d y}{d x}+4 y \operatorname{cosec}^2 x=0$. Changing the independent variable $x$ to $z$ by the substitution $z=\log \tan \frac{x}{2}$, the equation is changed to

Vedclass Test Series

Exam Paper Generator

Online Exam Module