SATHEE: Nuclear Chemistry - Result Question 4

Nuclear Chemistry - Result Question 4

####6. A human body required $0.01 M$ activity of radioactive substance after 24 hours. Half life of radioactive substance is 6 hours. Then injection of maximum activity of radioactive substance that can be injected will be

[2001]

(a) $0.08 M$

(b) $0.04 M$

(d) $0.16 M$

If species $_{a}^{b} X$ emits firstly a positron, then two $α$ and two $β$ and in last one $α$ and finally converted to species $_{c}^{d} Y$ , so correct relation is

[2001]

(a) $c = a - 5, d = b - 12$

(b) $c = a - 6, d = b - 8$

(d) $c = a - 5, d = b - 8$

Show Answer

Solution:

(d) Remaining activity $= 0.01 M$

after $24 h r s$

Remaining activity $=$ Initial activity $\times (\frac{1}{2})^{n}$ Used half life time $(n) = \frac{Total time}{T_{1 / 2}} = \frac{24}{6} = 4$

So, $0.01 =$ Initial activity $\times (\frac{1}{2})^{4}$

Initial activity $= 0.01 \times 16 = 0.16 M$

Nuclear Chemistry - Result Question 4

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Nuclear Chemistry - Result Question 4

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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