wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Consider the following nuclear reactions.
I. 147N+42He178O+X
II. 94Be+42H126He+Y. Then.

A
X and Y are both protons
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
X and Y are both neutrons
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
X is a proton and Y is a neutron
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
X is a neutron and Y is a proton
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C X is a proton and Y is a neutron
In the first reaction, no of protons on LHS is 7+2=9; the no protons on RHS is 8
Mass number on LHS is 14+4=18
Mass number on RHS is 17
Hence, the no of protons X will have is 98=1 and the no of neutrons X
has is 1817=1
In the 2nd reaction, no of protons Y has is 4+26=0 and no of neutron Y has is 9+412=1
Hence, both X and Y have one neutron.
In case of X energy can be released equivalent of one neutron mass.
Hence, X is a proton and Y is a neutron.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon