Question

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Answer 1

Answer:

$\sf Correct \: answer \: is \: option \: C \rightarrow\: 5\: grams$

Explanation:

Given:

Half life of cesium : 2 years

Amount of substance initial (N0)= 10 grams

To find:

Amount of substance remaining after 2 years(Nt) = ?

Solution:

Half life: The time interval in which the amount of substance at initial reduces to exact half of it, this reduction is seen in radioactive element.

Given by the formula,

$N_t = N_0 {( \frac{1}{2} )}^{ \frac{t}{ {t_{ \frac{1}{2}} } } }$

Where Nt is the amount of substance remaining after time t,

N0 is the amount of substance at time t = 0,

t is the time at which we have to find out how much substance disintegrated from t = 0 to t = t

& t_1/2 is the half life corresponding radioactive sample.

Now the question directly asks for the amount of substance remaining at half life and the initial amount is given, so we can directly half the initial amount, and the final answer would be

$\sf \frac{10}{2} = {\cancel\frac{10}{2}} \: ^{5} = 5 \: grams$

let's verify the above answer by calculating it with formula method, substituting the corresponding given data in above formula.

$N_t = 10 {( \frac{1}{2} )}^ \frac{2 \: years}{2 \: years} \\ \small \sf \:2 \: years \: and \: 2 \: years \: in \: power \: \\\sf \small \: has \: been \: cancelled \: out \\ \: N_t = 10 {( \frac{1}{2} )}^1 \\ N_t = \frac{10}{2} \\ \: \displaystyle \boxed{N_t = 5 \: grams}$

$\sf \small \: Answer \rightarrow 5 \: grams \: of \: sample \: would \: remain.$

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