SATHEE: Nuclear Chemistry - Result Question 1

Nuclear Chemistry - Result Question 1

####1. The half life of a substance in a certain enzyme catalysed reaction is $138 s$ . The time required fo the concentration of the substance to fall fron $1.28 m g L^{- 1}$ to $0.04 m g L^{- 1}$ , is :

[2011]

(a) $414 s$

(b) $552 s$

(d) $276 s$

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Solution:

Total time $T =$ no. of half lives $(n) \times$ half life $(t_{1 / 2})$

$\frac{N}{N_{0}} = (\frac{1}{2})^{n}$

where $n =$ no. of half lives

Give $N_{0}$ (original amount) $= 1.28 m g / l$

$N$ (amount of substance left after time $T$ )

$= 0.04 m g L^{- 1}$

$\begin{aligned} ∴ \frac{0.04}{1.28} & = (\frac{1}{2})^{n}; \frac{1}{32} = (\frac{1}{2})^{n}; (\frac{1}{2})^{5} = (\frac{1}{2})^{n} \\ n & = 5 \\ T & = 5 \times 138 = 690 \end{aligned}$

Nuclear Chemistry - Result Question 1

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Nuclear Chemistry - Result Question 1

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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