Question:
After two hours, one-sixteenth of the starting amount of a certain radioactive isotope remained undecayed. The half life of the isotope is
After two hours, one-sixteenth of the starting amount of a certain radioactive isotope remained undecayed. The half life of the isotope is
Updated On: Oct 25, 2024
- 15 minutes
- 30 minutes
- 45 minutes
- 4 hour
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
The correct option is (B): 30 minutes
\(N=N_{0}\left(\frac{1}{2}\right)^{n}\)
or \(\frac{N_{0}}{16}=N_{0}\left(\frac{1}{2}\right)^{n}\)
or \(n=4\)
Halflife \(t_{1/2}=\frac{t}{n}=\frac{2}{4}=\frac{1}{2} h\)
T=0.5hour=30 minutes.
\(N=N_{0}\left(\frac{1}{2}\right)^{n}\)
or \(\frac{N_{0}}{16}=N_{0}\left(\frac{1}{2}\right)^{n}\)
or \(n=4\)
Halflife \(t_{1/2}=\frac{t}{n}=\frac{2}{4}=\frac{1}{2} h\)
T=0.5hour=30 minutes.
Was this answer helpful?
1
2
Top Questions on Half life
- The bond angles H-O-N and O-N-O in the planar structure of nitric acid molecule are respectively:
- Observe the following reaction. This reaction represents: \[ 4HCl + O_2 \xrightarrow{CuCl_2, 723K} 2Cl_2 + 2H_2O \]
- A wooden plant accumulates $10 𝑚𝑔 𝑘𝑔^{−1}$ of $^{14}C$ during its life span. A fossil of this plant was discovered and contains $2.5 𝑚𝑔 𝑘𝑔^{−1}$ of $^{14}C$. The age of this fossil at the time of discovery is years (rounded off to the nearest integer).
(Use 5730 years as half-life of $^{14}C$) - Half life of a first order reaction is \(36\;hr\). Find out time (in hour) required for concentration of reactant to get reduced by \(90\%\).
- The number of incorrect statement/s from the following is ____ .
A. The successive half lives of zero order reactions decreases with time.
B. A substance appearing as reactant in the chemical equation may not affect the rate of reaction.
C. Order and molecularity of a chemical reaction can be a fractional number.
D. The rate constant units of zero and second order reaction are mol L–1s–1 and mol–1 Ls–1 respectively.
View More Questions
Questions Asked in BITSAT exam
- Consider the following two propositions: $$ P_1: \neg (p \rightarrow \neg q) $$ $$ P_2: (p \wedge \neg q) \wedge ((\neg p) \vee q) $$ If the proposition $p \rightarrow ((\neg p) \vee q)$ is evaluated as FALSE, then:
- BITSAT - 2024
- Statements
In the given cycle ABCDA, the heat required for an ideal monoatomic gas will be:
- BITSAT - 2024
- specific heat capacity
- In normal adjustment, for a refracting telescope, the distance between the objective and eyepiece is 30 cm. The focal length of the objective, when the angular magnification of the telescope is 2, will be:
- BITSAT - 2024
- thermal properties of matter
- If the distance between an object and its two times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be:
- BITSAT - 2024
- thermal properties of matter
- Young's double slit experiment is performed in a medium of refractive index \( 1.33 \). The maximum intensity is \( I_0 \). The intensity at a point on the screen where the path difference between the light coming out from slits is \( \lambda/4 \), is:
- BITSAT - 2024
- Dual nature of radiation and matter
View More Questions
Concepts Used:
Half-life
The amount of time taken for half of a particular sample to react is known as Half-life.
Half-Life Formula:
We can describe exponential decay by any of the three formulas
