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Chemical Kinetics and Nuclear Chemistry question

2004 · Shift 0 · Q43
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Chemical Kinetics and Nuclear Chemistry question

2004 · Shift 0 · Q43

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
The half – life of a radioisotope is four hours. If the initial mass of the isotope was 200 g, the mass remaining after 24 hours undecayed is
  1. A
    1.042 g
  2. B
    4.167 g
  3. C
    3.125 g
  4. D
    2.084 g
View written solutionFree

Correct answer: C

  1. Use the radioactive decay relation in terms of half-life

If the half-life is t1/2t_{1/2}t1/2​, then after time ttt, the remaining mass is

m=m0(12)t/t1/2m = m_0\left(\frac{1}{2}\right)^{t/t_{1/2}}m=m0​(21​)t/t1/2​

where:

  • m0=200 gm_0 = 200\,\text{g}m0​=200g
  • t1/2=4 ht_{1/2} = 4\,\text{h}t1/2​=4h
  • t=24 ht = 24\,\text{h}t=24h
  1. Find the number of half-lives elapsed

n=tt1/2=244=6n = \frac{t}{t_{1/2}} = \frac{24}{4} = 6n=t1/2​t​=424​=6

So, 6 half-lives have passed.

  1. Calculate the remaining mass

m=200(12)6m = 200\left(\frac{1}{2}\right)^6m=200(21​)6

Since

(12)6=164\left(\frac{1}{2}\right)^6 = \frac{1}{64}(21​)6=641​

therefore,

m=200×164=3.125 gm = 200 \times \frac{1}{64} = 3.125\,\text{g}m=200×641​=3.125g

  1. Match with the options
  • A: 1.042 g1.042\,\text{g}1.042g
  • B: 4.167 g4.167\,\text{g}4.167g
  • C: 3.125 g3.125\,\text{g}3.125g
  • D: 2.084 g2.084\,\text{g}2.084g

Hence, the correct option is:

C\boxed{\text{C}}C​

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