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The β − β activity of a sample of C O 2 C prepared form a contemporary wood gave a count rate of 25.5 25.5 counts per minute ( c p m ) ( . The same of C O 2 C form an ancient wooden statue gave a count rate of 20.5 c p m 20.5 , in the same counter condition. Calculate its age to the nearest 50 year taking t 1 / 2 t for . 14 C . as 5770 year. What would be the expected count rate of an identical mass of C O 2 C form a sample which is 4000 year old? Select the correct answer from above options

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Correct Answer - a . t = 821 y e a r s a , b . b Coutn rate = 15.5 c ± = The above magnitudes are same which means that the stability of all three atoms is same. Initial activity ( N 0 = 25.5 c ± ) ( Activity after some time ( N 1 ( or N ) = 20.5 c ± N a . a t = 2.303 K log t N 0 N t N = 2.303 × 5770 y e a r 0.69 log = 25.5 20.5 25.5 = 19258.4 × log 1.2439 = = 19258.4 × 0.0947 = 1821 y e a r s = b . b Number of counts decreases in 1821 years = 25.5 − 20.5 = 5 c ± = Number of counts of decrease in 4000 years 5 1821 × 4000 = 10 c o u n t s ≈ 10 c o u n t s 5 Count rate of a sample which is 4000 years old = 25.5 − 10 = 15.5 c ± =

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