The number of disintegrations per second that will occur in $1$ gram of $_{92}^{238} U$ is:
(if its half-life against $α$ decay is $1.42 \times 10^{17} s$ )
$1.2 \times 10^{5} s^{- 1}$
$2.1 \times 10^{5} s^{- 1}$
$1.5 \times 10^{5} s^{- 1}$
$0.1235 \times 10^{5} s^{- 1}$

Question

The number of disintegrations per second that will occur in $1$ gram of $_{92}^{238} U$ is:
(if its half-life against $α$ decay is $1.42 \times 10^{17} s$ )
$1.2 \times 10^{5} s^{- 1}$
$2.1 \times 10^{5} s^{- 1}$
$1.5 \times 10^{5} s^{- 1}$
$0.1235 \times 10^{5} s^{- 1}$

Toppr Admin · Accepted Answer

T1-2-1-42-xD7-1017s1-xA0-gm of U contains -6-xD7-1023235-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -25-3-xD7-1020 atomsThe decay rate-R-x3BB-N-xA0- -ln-xA0-2t1-2-xD7-25-3-xD7-1020-xA0- -xA0- -6931-42-xD7-1017-xD7-25-3-xD7-1020

The number of disintegrations per second that will occur in 1 gram of 23892U is:(if its half-life against α decay is 1.42×1017s)1.2×105s−12.1×105s−11.5×105s−10.1235×105s−1

t1/2=1.42×1017s1 gm of U contains =6×1023235 =25.3×1020 atomsThe decay rate:R=λN =ln 2t1/2×25.3×1020 =.6931.42×1017×25.3×1020

The number of disintegrations per second that will occur in $1$ gram of $_{92}^{238} U$ is:
(if its half-life against $α$ decay is $1.42 \times 10^{17} s$ )
$1.2 \times 10^{5} s^{- 1}$
$2.1 \times 10^{5} s^{- 1}$
$1.5 \times 10^{5} s^{- 1}$
$0.1235 \times 10^{5} s^{- 1}$

$t_{1 / 2} = 1.42 \times 10^{17} s$
$1 g m$ of U contains $= \frac{6 \times 10^{23}}{235}$
$= 25.3 \times 10^{20}$ atoms
The decay rate:
$R = λ N$
$= \frac{l n 2}{t_{1 / 2}} \times 25.3 \times 10^{20}$
$= \frac{.693}{1.42 \times 10^{17}} \times 25.3 \times 10^{20}$