Geiger counter reading of a radioactive sample is initially 6800 counts per minute. The same sample gives a reading of 425 counts per minute 10 h later. The half life of sample is α hours. Find 2 α.

Question

Geiger counter reading of a radioactive sample is initially 6800 counts per minute. The same sample gives a reading of 425 counts per minute 10 h later. The half life of sample is α hours. Find 2 α. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

A=(A0/2n) ⇒ 2n=(6800/425) (n= Number of half life) 2n=16=24 ⇒ n=4 ⇒ 4 × t1 / 2=10 ⇒ t1 / 2=2.5 h

Q. Geiger counter reading of a radioactive sample is initially $6800$ counts per minute. The same sample gives a reading of $425$ counts per minute $10 h$ later. The half life of sample is $α$ hours. Find $2 α$ .

$A = \frac{A _{0}}{2 ^{n}} \Rightarrow 2^{n} = \frac{6800}{425}$
( $n =$ Number of half life)
$2^{n} = 16 = 2^{4}$
$\Rightarrow n = 4$
$\Rightarrow 4 \times t_{1/2} = 10$
$\Rightarrow t_{1/2} = 2.5 h$

Q. Geiger counter reading of a radioactive sample is initially 6800 counts per minute. The same sample gives a reading of 425 counts per minute 10h later. The half life of sample is α hours. Find 2α.

A=2nA0​​⇒2n=4256800​ (n= Number of half life) 2n=16=24 ⇒n=4 ⇒4×t1/2​=10 ⇒t1/2​=2.5h

Q. Geiger counter reading of a radioactive sample is initially $6800$ counts per minute. The same sample gives a reading of $425$ counts per minute $10 h$ later. The half life of sample is $α$ hours. Find $2 α$ .

$A = \frac{A _{0}}{2 ^{n}} \Rightarrow 2^{n} = \frac{6800}{425}$
( $n =$ Number of half life)
$2^{n} = 16 = 2^{4}$
$\Rightarrow n = 4$
$\Rightarrow 4 \times t_{1/2} = 10$
$\Rightarrow t_{1/2} = 2.5 h$