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Atoms and Nuclei Part 1

Section: Physics

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Question : 1 of 100

Marks: +1, -0

Two radioactive substances

X

Y

originally have

N_{1}

N_{2}

nuclei, respectively. Half-life of

X

is half of the half-life of

Y

. After three half-lives of

Y

, number of nuclei of both are equal. The ratio

\;\frac{N_{1}}{N_{2}}

will be equal to

[25 Feb 2021 Shift 1]

$\;\frac{1}{8}$
$\;\frac{3}{1}$
$\;\frac{8}{1}$
$\;\frac{1}{3}$

Solution:

Given, initial amount of

X

and

Y

be

N_{1}

and

N_{2}

.

Let half-life of

X

be

t_{x}

and

y

be

t_{y}

.

According to question,

t_{x} = \;\frac{t_{y}}{2} = t

\Rightarrow \;\; t_{x}=t

and

t_{y}=2 t

After 3 half-lives of

Y

,

3 t_{y}=6 t

As we know that,

N=N_{0} e^{-\lambda t}

where,

N

is the number of nuclei left undecayed.

and

t_{1/2} \;\; = \;\frac{0.693}{\lambda}

\lambda \;\; = \;\frac{0.693}{t_{1/2}}

\lambda_{1} \;\; = \;\frac{0.693}{t} \;\text{and}\; \lambda_{2}=\;\frac{0.693}{2t}

Since, after 3 half-lives of

Y

number of nuclei of both become equal.

\therefore

\therefore \; N_{1} e^{-\lambda_{1} 6 t}=N_{2} e^{-\lambda_{2} 6 t}

\Rightarrow\; \frac{N_{1}}{N_{2}} = e^{6 t(-\lambda_{2}+\lambda_{1})}

\Rightarrow\; \frac{N_{1}}{N_{2}} = e^{6 t\left(\;\frac{0.693}{t}-\;\frac{0.693}{2t}\right)}

\;\; = e^{0.693\left(\;\frac{6t}{t}-\;\frac{6t}{2t}\right)} = e^{0.693 \times 3} = 7.9 \approx 8

\;\frac{N_{1}}{N_{2}} \;\; = \;\frac{8}{1}

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