(C) The nuclear reaction is given by: $_{13}Al^{28} + _{z}X^{a} \to _{15}P^{31} + _{0}n^{1}$ By balancing the atomic number: $13 + z = 15 + 0 \implies

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

(C) The nuclear reaction is given by: $_{13}Al^{28} + _{z}X^{a} \to _{15}P^{31} + _{0}n^{1}$ By balancing the atomic number: $13 + z = 15 + 0 \implies z = 2$. By balancing the mass number: $28 + a = 31 + 1 \implies a = 4$. The projectile with atomic number $2$ and mass number $4$ is an alpha particle $(_{2}He^{4})$.

Class 12 Chemistry Nuclear Chemistry Artificial transmutation

Medium

$_{13}Al^{28}$ when radiated by a suitable projectile gives $_{15}P^{31}$ and a neutron. The projectile used is:

A
Proton
B
Neutron
C
Alpha particle
D
Deuteron

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Nuclear Chemistry

Subtopic

Artificial transmutation

In the nuclear reaction $_{92}U^{238} \to _{82}Pb^{206} + x\,_{2}He^{4} + y\,_{-1}\beta^{0}$,the values of $x$ and $y$ are respectively......

Medium

View Solution

$_{Z}X^{M} + _{2}He^{4} \to _{15}P^{30} + _{0}n^{1}$. Then

Medium

View Solution

For the fission reaction $_{92}U^{235} + _0n^1 \to _{56}Ba^{140} + _yE^x + 2_0n^1$,the values of $x$ and $y$ are:

Medium

View Solution

In the nuclear reaction $_{12}Mg^{24} + _{2}He^{4} = _{0}n^{1} + ?$,the product nucleus is:

Medium

View Solution

The nuclear reaction $_{17}Cl^{37} + _1H^2 \to _{18}Ar^{38} + _0n^1$ represents:

Medium

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

$_{13}Al^{28}$ when radiated by a suitable projectile gives $_{15}P^{31}$ and a neutron. The projectile used is:

Similar Questions

In the nuclear reaction $_{92}U^{238} \to _{82}Pb^{206} + x\,_{2}He^{4} + y\,_{-1}\beta^{0}$,the values of $x$ and $y$ are respectively......

$_{Z}X^{M} + _{2}He^{4} \to _{15}P^{30} + _{0}n^{1}$. Then

For the fission reaction $_{92}U^{235} + _0n^1 \to _{56}Ba^{140} + _yE^x + 2_0n^1$,the values of $x$ and $y$ are:

In the nuclear reaction $_{12}Mg^{24} + _{2}He^{4} = _{0}n^{1} + ?$,the product nucleus is:

The nuclear reaction $_{17}Cl^{37} + _1H^2 \to _{18}Ar^{38} + _0n^1$ represents:

Vedclass Test Series

Exam Paper Generator

Online Exam Module