Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
78 views
in Physics by (91.3k points)
closed by
8.0 m Ci सक्रियता का रेडियोऐक्टिव स्त्रोत प्राप्त करने के लिए `""_(27)^(60)"Co"` की कितनी मात्रा की आवश्यकता होगी ? `""_(27)^(60)" Co"` की अर्ध - आयु 5.3 वर्ष है |

1 Answer

0 votes
by (90.7k points)
selected by
 
Best answer
दिया है - `R=8.0" m Ci"`
`=8.0xx10^(-3)xx3.7xx10^(10)`
disintegration per second
`=29.6xx10^(7)`
`=2.96xx10^(8)` disintegration per second
`T_(1//2)=5.3` वर्ष
`=5.3xx365xx24xx60xx60" s"`
`=1.67xx10^(8)" s"`
`:. lambda=(0.6931)/(T_(1//2))=(0.6931)/(1.67xx10^(8))=0.415xx10^(-8)`
`=4.15xx10^(-9)s^(-1)`
सूत्र - `R=lambdaN` से,
`N=(R)/(lambda)=(2.97xx10^(8))/(4.15xx10^(-9))=0.7157xx10^(17)`
`=7.157xx10^(16)`
Co के एक मोल में अर्थात 60 ग्राम में `6.023xx10^(23)` परमाणु होते हैं |
अत: उसके `7.157xx10^(15)` परमाणु का द्रव्यमान
`=(60)/(6.023xx10^(23))xx7.157xx10^(16)`
`=71.30xx10^(-7)" g"`
`=7.13xx10^(-6)" g."`

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...