(A) Cobalt-$60$ $(^{60}Co)$ is a radioactive isotope of cobalt.It undergoes radioactive decay and emits high-energy $\gamma$-radiations.These $\gamma$

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

If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

(A) Cobalt-$60$ $(^{60}Co)$ is a radioactive isotope of cobalt.It undergoes radioactive decay and emits high-energy $\gamma$-radiations.These $\gamma$-radiations have high penetrating power and are used in radiation therapy to target and destroy cancerous cells.Therefore,the Assertion is correct,the Reason is correct,and the Reason provides the correct explanation for the Assertion.

Class 12 Physics Nuclei Properties of Alpha, Beta and Gamma Rays and Decay Process

Easy

Assertion: Cobalt-$60$ is useful in cancer therapy.
Reason: Cobalt-$60$ is a source of $\gamma$-radiations capable of killing cancerous cells.

AIIMS 2006 All AIIMS

A
If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
B
If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
C
If the Assertion is correct but Reason is incorrect.
D
If both the Assertion and Reason are incorrect.

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Nuclei

Subtopic

Properties of Alpha, Beta and Gamma Rays and Decay Process

Previous Year

AIIMS 2006 Paper All AIIMS years →

In the nuclear reaction ${}_{92}^{235}U$ decaying to ${}_{91}^{231}Pa$,what are the particles emitted?

Difficult

View Solution

$A$ radioactive reaction is $_{92}U^{238} \to _{82}Pb^{206}$. How many $\alpha$ and $\beta$ particles are emitted?

Medium

View Solution

In the given nuclear reaction,$A, B, C, D, E$ represent:
$_{92}U^{238} \xrightarrow{\alpha} _{B}Th^{A} \xrightarrow{\beta} _{D}Pa^{C} \xrightarrow{E} _{92}U^{234}$

Medium

View Solution

Find the $Q$-value and the kinetic energy of the emitted $\alpha$-particle in the $\alpha$-decay of $(a) \; ^{226}_{88} Ra$ and $(b) \; ^{220}_{86} Rn$.
Given:
$m(^{226}_{88} Ra) = 226.02540 \; u$
$m(^{222}_{86} Rn) = 222.01750 \; u$
$m(^{220}_{86} Rn) = 220.01137 \; u$
$m(^{216}_{84} Po) = 216.00189 \; u$
$m(^{4}_{2} He) = 4.002603 \; u$

Medium

View Solution

$A$ nucleus $X$ undergoes the following transformation:
$X \xrightarrow{\alpha} Y$
$Y \xrightarrow{2\beta} Z$
Then:

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Assertion: Cobalt-$60$ is useful in cancer therapy.Reason: Cobalt-$60$ is a source of $\gamma$-radiations capable of killing cancerous cells.

Similar Questions

In the nuclear reaction ${}_{92}^{235}U$ decaying to ${}_{91}^{231}Pa$,what are the particles emitted?

$A$ radioactive reaction is $_{92}U^{238} \to _{82}Pb^{206}$. How many $\alpha$ and $\beta$ particles are emitted?

In the given nuclear reaction,$A, B, C, D, E$ represent:$_{92}U^{238} \xrightarrow{\alpha} _{B}Th^{A} \xrightarrow{\beta} _{D}Pa^{C} \xrightarrow{E} _{92}U^{234}$

$A$ nucleus $X$ undergoes the following transformation:$X \xrightarrow{\alpha} Y$$Y \xrightarrow{2\beta} Z$Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion: Cobalt-$60$ is useful in cancer therapy.
Reason: Cobalt-$60$ is a source of $\gamma$-radiations capable of killing cancerous cells.

In the given nuclear reaction,$A, B, C, D, E$ represent:
$_{92}U^{238} \xrightarrow{\alpha} _{B}Th^{A} \xrightarrow{\beta} _{D}Pa^{C} \xrightarrow{E} _{92}U^{234}$

$A$ nucleus $X$ undergoes the following transformation:
$X \xrightarrow{\alpha} Y$
$Y \xrightarrow{2\beta} Z$
Then: