
Which of the following can be used to convert $_{7}^{14}N$ into $_{8}^{17}O$ ?
(A) Deuterons
(B) protons
(C) α-particles
(D) Neutron
Answer
163.5k+ views
Hint: We can solve this question by checking the atomic mass of each of the particles given in the options. A particle with a mass less than the atomic mass change from nitrogen (N) to oxygen (O) cannot be used to convert$_{7}^{14}N$ into$_{8}^{17}O$.
Complete Step by Step Solution:
During conversion of $_{7}^{14}N$ into $_{8}^{17}O$ , the atomic mass changes by
$\Delta m=17-14=3$.
This means that the particle required to convert $_{7}^{14}N$ into $_{8}^{17}O$ should have a mass greater than or equal to 3.
(A) Deuterons ${{(}_{1}}{{H}^{2}})$ have a mass of 2, which is less than 3.$(2<3)$
So, deuteron cannot be used to convert $_{7}^{14}N$ into$_{8}^{17}O$.
(B) Protons ${{(}_{1}}{{H}^{1}})$ have a mass of 1, which is less than 3.$(1<3)$
So, protons cannot be used to convert $_{7}^{14}N$ into$_{8}^{17}O$.
(C) α-particles ${{(}_{2}}H{{e}^{4}})$ have a mass of 4, which is more than 3.\[(4>3)\]
$_{7}{{N}^{14}}{{+}_{2}}H{{e}^{4}}{{\xrightarrow{{}}}_{8}}{{O}^{17}}{{+}_{1}}{{H}^{1}}$
So, α-particles can be used to convert $_{7}^{14}N$ into$_{8}^{17}O$.
(D) Neutron ${{(}_{0}}{{n}^{1}})$ has a mass of 1, which is less than 3.$(1<3)$
So, neutron cannot be used to convert $_{7}^{14}N$ into$_{8}^{17}O$.
Correct Option: (C) α-particles .
Additional Information: Alpha particles are made up of 2 protons and 2 neutrons. The protons and neutrons in the alpha particle are held together by a strong nuclear force. It is a positively charged particle. These have very low penetrating power, such that they cannot even penetrate the outer dead layer of skin. The alpha particles are very large and move very slowly. This is why it can be stopped completely by a very thin sheet of paper.
Note: The mass of each of the particle, given in the options should be correctly known to give the correct answer to this question.
Complete Step by Step Solution:
During conversion of $_{7}^{14}N$ into $_{8}^{17}O$ , the atomic mass changes by
$\Delta m=17-14=3$.
This means that the particle required to convert $_{7}^{14}N$ into $_{8}^{17}O$ should have a mass greater than or equal to 3.
(A) Deuterons ${{(}_{1}}{{H}^{2}})$ have a mass of 2, which is less than 3.$(2<3)$
So, deuteron cannot be used to convert $_{7}^{14}N$ into$_{8}^{17}O$.
(B) Protons ${{(}_{1}}{{H}^{1}})$ have a mass of 1, which is less than 3.$(1<3)$
So, protons cannot be used to convert $_{7}^{14}N$ into$_{8}^{17}O$.
(C) α-particles ${{(}_{2}}H{{e}^{4}})$ have a mass of 4, which is more than 3.\[(4>3)\]
$_{7}{{N}^{14}}{{+}_{2}}H{{e}^{4}}{{\xrightarrow{{}}}_{8}}{{O}^{17}}{{+}_{1}}{{H}^{1}}$
So, α-particles can be used to convert $_{7}^{14}N$ into$_{8}^{17}O$.
(D) Neutron ${{(}_{0}}{{n}^{1}})$ has a mass of 1, which is less than 3.$(1<3)$
So, neutron cannot be used to convert $_{7}^{14}N$ into$_{8}^{17}O$.
Correct Option: (C) α-particles .
Additional Information: Alpha particles are made up of 2 protons and 2 neutrons. The protons and neutrons in the alpha particle are held together by a strong nuclear force. It is a positively charged particle. These have very low penetrating power, such that they cannot even penetrate the outer dead layer of skin. The alpha particles are very large and move very slowly. This is why it can be stopped completely by a very thin sheet of paper.
Note: The mass of each of the particle, given in the options should be correctly known to give the correct answer to this question.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Solutions Class 12 Notes: CBSE Chemistry Chapter 1

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

Electrochemistry Class 12 Notes: CBSE Chemistry Chapter 2
