Chemical Engineering - Nuclear Power Engineering - Discussion
Discussion Forum : Nuclear Power Engineering - Section 1 (Q.No. 15)
15.
The amount of a radioisotope remaining undecayed after a time equal to four times its half life, will be __________ percent.
Discussion:
3 comments Page 1 of 1.
Amal said:
4 years ago
@Harendra Singh.
How this equation came? Explain.
How this equation came? Explain.
Amal said:
4 years ago
Please explain.
Harendra Singh said:
7 years ago
t1/2 = (.693/K).
ln(Ca/Ca0) = -K * t.
Using both equations, and put t=4 * t1/2.
Ca/Ca0=e^(-K*4*.693/K) = 0.0625 or 6.25%.
THANKS
ln(Ca/Ca0) = -K * t.
Using both equations, and put t=4 * t1/2.
Ca/Ca0=e^(-K*4*.693/K) = 0.0625 or 6.25%.
THANKS
(2)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers