A radioactive element 90X238 decays into 83Y222. The number of β - particles emitted is :
Nikhila Thakur , 8 Years ago
Grade 12th pass
Modern Physics> A radioactive element 90X238 decays into ...
1 Answers
Khimraj
Last Activity: 8 Years ago
First of all calculate no of alpha particle emitted by particle. Which can be easily calculated by mass difference and are 238-222/4 = 4 so after alpha emotion element will be 82X222. So from here we easily can see there is only one beta particle is emitted as atomic no increase by one. So correct option is (4). Hope it clears. Please if u like answer then approve the answer.
Provide a better Answer & Earn Cool Goodies
LIVE ONLINE CLASSES
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Ask a Doubt
Get your questions answered by the expert for free
Other Related Questions on modern physics
Two particles approach each other from very large distance as shown, . if the least separation between the particles is b, what is the speed of M in center of mass frame when particles are closest? (gravity negligible)
Two particles approach each other from very large distance as shown, . if the least separation between the particles is b, what is the speed of M in center of mass frame when particles are closest? (gravity negligible)
modern physics
0 Answer Available
Last Activity: 2 Years ago
Can the sum of two equal vectors be equal to either vector
Can the sum of two equal vectors be equal to either vector
modern physics
0 Answer Available
Last Activity: 2 Years ago
A nucleus X184 at rest emitt alpha particle With kinetic energy 9.0MeV. The product Nucleus is Y. The values of Binding energy per nucleon of alpha particle & Y differ by 1.5MeV. The values of Binding energy per nucleon of X & Y differ by (n÷230) MeV, then n is
A nucleus X184 at rest emitt alpha particle With kinetic energy 9.0MeV. The product Nucleus is Y. The values of Binding energy per nucleon of alpha particle & Y differ by 1.5MeV. The values of Binding energy per nucleon of X & Y differ by (n÷230) MeV, then n is
modern physics
0 Answer Available
Last Activity: 3 Years ago
SUBJECT: PROBLEM BASED ON LINE DETECTION.
Query: How to apply all the filters for line detection on the following piece of image (provided below)? Explain its Observations to comment on its results.
SUBJECT: PROBLEM BASED ON LINE DETECTION.
Query: How to apply all the filters for line detection on the following piece of image (provided below)? Explain its Observations to comment on its results.
Query: How to apply all the filters for line detection on the following piece of image (provided below)? Explain its Observations to comment on its results.
modern physics
0 Answer Available
Last Activity: 3 Years ago