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Hint: First order reaction: A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.

The rate increases as the number of times the concentration of reaction is increased. The half-life of first-order reactions: Half-life of a reaction is the time taken for half of the reaction to be completed.

Now,

${ t }_{ 1/2 }{ =ln2/k }$………..(1)

where , ${ t }_{ 1/2 }$ = half life

k= reaction rate constant

ln2 = base e logarithm

It is given that,

${ 75\% }$ of the active nuclei present in the sample initially will decay in time, so, ${ 25\% }$ is left undecayed.

${ 75\% }$ consumption = ${ 25\% }$ nuclei remains

Using the formula; ${ N= }{ N }_{ 0 }{ e }^{ -\lambda t }$……..(2)

where, ${ N }_{ 0 }$ = number of radioactive nuclei at any arbitrary time, ${ t }_{ 0 }$

N = number of radioactive nuclei at any time,t

${ \lambda }$ = decay constant

t = time

Let, initially decay = ${ N }_{ 0 }$

After ${ 75\% }$ decay = ${ N }_{ 0 }{ \div 4 }$

By putting the values in equation (2), we get

${ N }_{ 0 }{ \div 4 }= { N }_{ 0 }{ e }^{ -\lambda t }$

${ 1\div 4{ =1\div { e }^{ \lambda t } } }$

${ e }^{ \lambda t }{ =4 }$

Taking ln on both the sides, we get

${ ln }_{ e }{ e }^{ \lambda t }{ =ln4 }$

${ \lambda t\quad =\quad ln{ 2 }^{ 2 } }$

${ \lambda t\quad =\quad 2ln2 }$ ……….(3)

Now, as we know that ${ \lambda =1\div T }$

where, ${ \lambda }$ = decay constant

T = mean life

By putting the value of ${ \lambda }$ in equation (3), we get

${ (1\div T)t\quad =2(ln2) }$

${ t\quad =2(ln2)T }$

Hence, the correct option is D.

Note: The possibility to make a mistake is that you may confuse between log and ln in the formulas. ${ ln=2.303log }$. Don’t forget to apply this formula.

The rate increases as the number of times the concentration of reaction is increased. The half-life of first-order reactions: Half-life of a reaction is the time taken for half of the reaction to be completed.

__Complete step-by-step solution:__Now,

${ t }_{ 1/2 }{ =ln2/k }$………..(1)

where , ${ t }_{ 1/2 }$ = half life

k= reaction rate constant

ln2 = base e logarithm

It is given that,

${ 75\% }$ of the active nuclei present in the sample initially will decay in time, so, ${ 25\% }$ is left undecayed.

${ 75\% }$ consumption = ${ 25\% }$ nuclei remains

Using the formula; ${ N= }{ N }_{ 0 }{ e }^{ -\lambda t }$……..(2)

where, ${ N }_{ 0 }$ = number of radioactive nuclei at any arbitrary time, ${ t }_{ 0 }$

N = number of radioactive nuclei at any time,t

${ \lambda }$ = decay constant

t = time

Let, initially decay = ${ N }_{ 0 }$

After ${ 75\% }$ decay = ${ N }_{ 0 }{ \div 4 }$

By putting the values in equation (2), we get

${ N }_{ 0 }{ \div 4 }= { N }_{ 0 }{ e }^{ -\lambda t }$

${ 1\div 4{ =1\div { e }^{ \lambda t } } }$

${ e }^{ \lambda t }{ =4 }$

Taking ln on both the sides, we get

${ ln }_{ e }{ e }^{ \lambda t }{ =ln4 }$

${ \lambda t\quad =\quad ln{ 2 }^{ 2 } }$

${ \lambda t\quad =\quad 2ln2 }$ ……….(3)

Now, as we know that ${ \lambda =1\div T }$

where, ${ \lambda }$ = decay constant

T = mean life

By putting the value of ${ \lambda }$ in equation (3), we get

${ (1\div T)t\quad =2(ln2) }$

${ t\quad =2(ln2)T }$

Hence, the correct option is D.

Note: The possibility to make a mistake is that you may confuse between log and ln in the formulas. ${ ln=2.303log }$. Don’t forget to apply this formula.

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