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GATE_2018

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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)

$ 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,
 
$\lambda =\frac{ln 2}{t_{1/2}}$

$\lambda =\frac{0.693}{t_{1/2}}$

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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9 comments

  1. Sir can write the all geophysical formulas and upload it will be helpful for gate or net exam .

    ReplyDelete
    Replies
    1. Ok....Please follow this blog, if you like it.

      Delete
    2. Yes sir,,please give all the formulas.it will be really helpful.

      Delete
  2. please solve question no. 72 gate 2004.

    ReplyDelete
    Replies
    1. I don't have 2004 GATE paper. Please share the question?

      Delete
    2. The 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 _
      1)5305 years 2)6350 years 3) 7305 years 4)7500 years
      the ans is option (3)

      Delete


    3. The 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

      Delete
    4. I was supposed to put half life 5730 yr but now i got it. Thanku

      Delete
    5. Above equation came from (N/N_0) = (1/2)^t/t_1/2


      Here substitute the C14 half life =5730 years, then we will get the above equation (almost equal).

      Delete

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