Assuming 1 μ g of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is × 10-1 μ g. [Given: ln 10=2.303 ; log 2=0.30]

Question

Assuming 1 μ g of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is × 10-1 μ g. [Given: ln 10=2.303 ; log 2=0.30] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

t =(1/λ) ln (( a / a - x )) 100=(30/ℓ n 2) ln ((1/ w )) (1/ w )=10 W =0.1 × μ g Ans. 1 × 10-1 μ g

Q. Assuming $1 μg$ of trace radioactive element $X$ with a half life of 30 years is absorbed by a growing tree. The amount of $X$ remaining in the tree after 100 years is ___ $\times 1 0^{- 1} μg$ .
[Given : $ln 10 = 2.303; lo g 2 = 0.30]$

$t = \frac{1}{λ} ln (\frac{a}{a - x})$
$100 = \frac{30}{ℓ n 2} ln (\frac{1}{w})$
$\frac{1}{w} = 10$
$W = 0.1 \times μg$
Ans. $1 \times 1 0^{- 1} μg$

Q. Assuming 1μg of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is ___×10−1μg. [Given : ln10=2.303;log2=0.30]

t=λ1​ln(a−xa​) 100=ℓn230​ln(w1​) w1​=10 W=0.1×μg Ans. 1×10−1μg

Q. Assuming $1 μg$ of trace radioactive element $X$ with a half life of 30 years is absorbed by a growing tree. The amount of $X$ remaining in the tree after 100 years is ___ $\times 1 0^{- 1} μg$ .
[Given : $ln 10 = 2.303; lo g 2 = 0.30]$

$t = \frac{1}{λ} ln (\frac{a}{a - x})$
$100 = \frac{30}{ℓ n 2} ln (\frac{1}{w})$
$\frac{1}{w} = 10$
$W = 0.1 \times μg$
Ans. $1 \times 1 0^{- 1} μg$