GATE_2018
80)
The ratio of the number of daughter nuclides to the number of parent
nuclides after a decay period of 3 half-lives is _______.
(Thanks to Chandra Sekhar, ANU)
Solution:
$D= P_0-P----(1)$
$P=P_0*e^{-\lambda t}$
$P_0=P*e^{\lambda t}---(2)$
Substitute the above eq(2) in eq(1)
$P=P_0*e^{-\lambda t}$
$P_0=P*e^{\lambda t}---(2)$
Substitute the above eq(2) in eq(1)
$ D = P(e^{\lambda t} -1)
$--------------------------(3)
Givent that $t= 3 t_{\frac{1}{2}}$
$\frac{D}{P}$ = ?
By substituting the given values in the equation (3)
$D = P(e^{\lambda t} -1)$
$\frac{D}{P}= e^{\lambda t} -1$
here,
$\lambda =\frac{ln 2}{t_{1/2}}$
$\lambda =\frac{0.693}{t_{1/2}}$
$\lambda =\frac{0.693}{t_{1/2}}$
Substitute the $\lambda$ value in above equation
=$e^{\frac{0.693}{t_{\frac{1}{2}}} \times
3t_{\frac{1}{2}} } -1$
=$e^{0.693\times 3} -1$
= $e^{2.079} -1$
=7.99 – 1
$\frac{D}{P} =6.99$
Reference : Fundamental of Geophysics, William Lowrie.
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Deleteplease solve question no. 72 gate 2004.
ReplyDeleteI don't have 2004 GATE paper. Please share the question?
DeleteThe specific radiocarbon activity of wood sample is 5.25 dpm/gm of carbon. The specific activity of NBS oxalic acid standard is 13.37 dpm/gm of carbon. The age of the wood sample will be _
Delete1)5305 years 2)6350 years 3) 7305 years 4)7500 years
the ans is option (3)
DeleteThe Conventional Radiocarbon Ages (BP) is calculated using the radiocarbon decay equation:
t = -8033 ln (Asn/Aon)
t- age
Asn =5.25
Aon = 13.37
Where -8033 represents the mean lifetime of 14 C (Stuiver and Polach, 1977). Aon is the activity in counts per minute of the modern standard. Asn is the equivalent cpm for the sample. 'ln' represents the natural logarithm.
t= -8033 ln (5.25 /13.37)
t= -8033 (-0.934)
age t = 7509 years
(dpm/gm = disintegration per minute /gram)
Reference: https://www.radiocarbon.com/PDF/AMS-Methodology.pdf
I was supposed to put half life 5730 yr but now i got it. Thanku
DeleteAbove equation came from (N/N_0) = (1/2)^t/t_1/2
DeleteHere substitute the C14 half life =5730 years, then we will get the above equation (almost equal).