CSIR -NET DEC 2014

CSIR – NET DEC – 2014

79) A 1000 years old tree and a 10 year old gastropod were buried during an earthquake. After 5730 years the specific activity of ¹⁴C would be

A) Ten times higher in the tree than in the gastropad.

B) Hundred times higher in the tree than in the gastropod.

C) the same in both

D) lower in the tree than in the gastropod.

(Thanks to Neeraja, AU)

Solution:

The relation b/w Half life and time

$\frac{N}{N_o}=(\frac{1}{2})^{\frac{t}{t_{1/2}}}---(1)$

N - Nuclei present after time ‘t’ in the sample

N_o – Original Nuclei present in the sample

t_1/2 – Half-life of Sample (¹⁴C – 5730 years)

t – time

t_1/2 = 5730

$\frac{N}{N_o}=(\frac{1}{2})^{\frac{1000}{5730}}=(\frac{1}{2})^{0.174}$

$(\frac{N}{N_o})_{tree}=0.886$

t_1/2=5730

$(\frac{N}{N_o})=(\frac{1}{2})^{\frac{10}{5730}}=(\frac{1}{2})^{0.00174}$

$(\frac{N}{N_o})_{gastopad}=0.998$

$\therefore$ lower in the tree than in the gastropod.

GEOPHYSICS