Q.a radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 years. The time, in years, after which one-fourth of the material remains is.
a). 1080 b). 2430
b). 3240 d). 4860
explain with clear explanation. {class 12th Q.}
[last 6 year continous asked question in board exam of 5mark.]
a). 1080 b). 2430
b). 3240 d). 4860
explain with clear explanation. {class 12th Q.}
[last 6 year continous asked question in board exam of 5mark.]
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since,from Rutherford _soddy low,
the number of atoms left half lives is given by,
N=Nnot (1/2)^n
where,No is the original number of atims
The number of half lives in=time of decay/egfective half-cycle
relation between effective distegration constant (lemda)&half life(T)
lemda=In2/T
lemda1+lemda2=In2/T1+In2/T2
effective half life,
1/T=1/T1+1/T2
=1/1620+1/810
1/T=1+2/1620
T=540 year
n=T/540
N=N not (1/2)^t/540
N/N not=(1/2)^2
=(1/2)^t/540
t/540=2
t=2×540=1080year
hope it helps u topper
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