Question

A radionuclide ⁵⁶Mn is being produced in cyclotron at constant rate P by bombarding a maganese target with deutrons, ⁵⁶Mn has a half life of 2.5 hours and the target contains large number of only the stable manganese isotope ⁵⁵Mn. The reaction that produces 56Mn is

⁵⁵Mn + d → 56Mn + p

After being bombardment for a long time, the activity of ⁵⁶Mn becomes constant & equal to

13.8 × 10¹⁰S^–1, (use ln2 = 0.693, Avagadros number = 6 × 10²³, Atomic weight of ⁵⁶Mn = ⁵⁶gm/mole

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Linked Question 1

At what constant rate P, ⁵⁶Mn nuclei are being produced in the cyclotron during the bombardment?

2 × 10¹¹ nuclei/s

9.6 × 10¹⁰ nuclei/s

6.93 × 10¹⁰ nuclei/s

13.86 × 10¹⁰ nuclei/s

No Solution

Linked Question 2

After the activity of ⁵⁶Mn becomes constant, number of ⁵⁶Mn nuclei present in the target, equal to

1.2 × 10¹⁴

20 × 10¹¹

1.8 × 10¹⁵

5 × 10¹¹

(a) In equalibrium

rate of decay = rate of production

(b) as rate of decay = rate of production

$P = λN \Rightarrow N = \frac{P}{λ} = \frac{{Pt}_{1 / 2}}{\ln 2}$

$= \frac{13 .86 \times 10^{10} \times 2 .5 \times 60 \times 60}{0 .693} = 1 .8 \times 10^{15}$