The half-life period of a radioactive element is 140 days. After 650 days, one gram of the element will reduce to
- $\frac{1}{2}g$
- $\frac{1}{4}g$
- $\frac{1}{8}g$
- $\frac{1}{16}g$
The Correct Option is D
Approach Solution - 1
Amount of reactant remaining after /z-half-lives
$=\bigg(\frac{1}{2}\bigg)^n\times $ initial amount $=\bigg(\frac{1}{2}\bigg)^4\times1.0g=\frac{1}{16}g$
Approach Solution -2
Let's correct and restate the calculation using the correct elapsed time of 560 days:
Given:
- Half-life period (\( t_{1/2} \)) = 140 days
- Elapsed time (\( t \)) = 560 days
- Initial quantity (\( N_0 \)) = 1 gram
First, determine the number of half-lives that have passed:
\[ \text{Number of half-lives} = \frac{560}{140} = 4 \]
Next, use the formula for the remaining amount after \( n \) half-lives:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^n \]
Where \( n \) is the number of half-lives. Substituting the values:
\[ N(560) = 1 \left( \frac{1}{2} \right)^4 \]
Calculate \( \left( \frac{1}{2} \right)^4 \):
\[ \left( \frac{1}{2} \right)^4 = \frac{1}{16} \]
Thus, the remaining quantity of the element after 560 days is:
\[ N(560) = \frac{1}{16} \, \text{grams} \]
Therefore, the amount of the element remaining after 560 days is:
\[ \boxed{\frac{1}{16} \, \text{grams}} \]
So Correct Answer is Option D :\(\frac{1}{16}gm\)
Top Questions on Chemical Kinetics
- The thermal decomposition of an acid is a first-order reaction with a rate constant of \(2.3 \times 10^{-5}\ \text{s}^{-1}\) at a certain temperature. Calculate how long it will take for three-fourths of the initial quantity of acid to decompose. (\(\log 4 = 0.6021\), \(\log 2 = 0.3010\))
- CBSE Compartment XII - 2025
- Chemistry
- Chemical Kinetics
- Half-life of zero-order reaction $ A \to $ product is 1 hour, when initial concentration of reaction is 2.0 mol L$^{-1}$. The time required to decrease concentration of A from 0.50 to 0.25 mol L$^{-1}$ is:
- JEE Main - 2025
- Chemistry
- Chemical Kinetics
- The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K. If the energy barrier with respect to reactant energy for such isomeric transformation is 191.48 kJ mol\(^{-1}\) and the frequency factor is \(10^{20}\), the time required for 50% molecules of A to become B is ________________________ picoseconds (nearest integer). [R = 8.314 J K\(^{-1}\) mol\(^{-1}\)]
- JEE Main - 2025
- Chemistry
- Chemical Kinetics
- Consider an elementary reaction: \[ A(g) + B(g) \rightarrow C(g) + D(g) \] If the volume of the reaction mixture is suddenly reduced to \( \frac{1}{3} \) of its initial volume, the reaction rate will become \( x \) times of the original reaction rate. The value of \( x \) is:
- JEE Main - 2025
- Chemistry
- Chemical Kinetics
- Drug X becomes ineffective after 50% decomposition. The original concentration of drug in a bottle was 16 mg/mL which becomes 4 mg/mL in 12 months. The expiry time of the drug in months is ____ . Assume that the decomposition of the drug follows first order kinetics.
- JEE Main - 2025
- Chemistry
- Chemical Kinetics
Questions Asked in JEE Advanced exam
- The total number of real solutions of the equation $$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $$ is
(Here, the inverse trigonometric functions $ \sin^{-1} x $ and $ \tan^{-1} x $ assume values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)- JEE Advanced - 2025
- Inverse Trigonometric Functions
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
- JEE Advanced - 2025
- Coordinate Geometry
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
- JEE Advanced - 2025
- Waves and Oscillations
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
- If $$ \alpha = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} x}{2x^2 - 3x + 2} \, dx, $$ then the value of $ \sqrt{7} \tan \left( \frac{2\alpha \sqrt{7}}{\pi} \right) $ is.
(Here, the inverse trigonometric function $ \tan^{-1} x $ assumes values in $ \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) $.)- JEE Advanced - 2025
- Integral Calculus
Concepts Used:
Chemical Kinetics
Chemical kinetics is the description of the rate of a chemical reaction. This is the rate at which the reactants are transformed into products. This may take place by abiotic or by biological systems, such as microbial metabolism.
Rate of a Chemical Reaction:
The speed of a reaction or the rate of a reaction can be defined as the change in concentration of a reactant or product in unit time. To be more specific, it can be expressed in terms of: (i) the rate of decrease in the concentration of any one of the reactants, or (ii) the rate of increase in concentration of any one of the products. Consider a hypothetical reaction, assuming that the volume of the system remains constant. R → P
Read More: Chemical Kinetics MCQ
Factors Affecting The Reaction Rate:
- The concentration of Reactants - According to collision theory, which is discussed later, reactant molecules collide with each other to form products.
- Nature of the Reactants - The reaction rate also depends on the types of substances that are reacting.
- Physical State of Reactants - The physical state of a reactant whether it is solid, liquid, or gas can greatly affect the rate of change.
- Surface Area of Reactants - When two or more reactants are in the same phase of fluid, their particles collide more often than when either or both are in the solid phase or when they are in a heterogeneous mixture. In a heterogeneous medium, the collision between the particles occurs at an interface between phases. Compared to the homogeneous case, the number of collisions between reactants per unit time is significantly reduced, and so is the reaction rate.
- Temperature - If the temperature is increased, the number of collisions between reactant molecules per second. Increases, thereby increasing the rate of the reaction.
- Effect Of Solvent - The nature of the solvent also depends on the reaction rate of the solute particles.
- Catalyst - Catalysts alter the rate of the reaction by changing the reaction mechanism.