What is steady state number of $I$ after a long time?

3.2 \times 10^{9}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.5 \times 10^{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.3 \times 10^{9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $3.2 \times 10^{9}$
$\frac{d N}{d t}$ = $N λ$
Given that: $t_{1 / 2} = 25 m i n = 1500 s$
$λ = \frac{l n (2)}{t_{1 / 2}} = \frac{0.693}{1500}$
$N$ $= \frac{d N / d t}{λ}$
$N$ $= \frac{1.5 \times 10^{6} \times 1500}{.693}$ $= 3.2 \times 10^{9}$

Suggest Corrections

Similar questions

Q. A

0.50 g

sample of rock was found to have

2.5 \times 10^{- 6} mol

_{19}^{40} K (t_{1 / 2} = 1.3 \times 10^{9} years)

and

7.5 \times 10^{- 6} {mol of}_{20}^{40} C a

. How old is the rock?

A radioactive sample of $U^{238}$ decay to $P b$ through a process for which half life is $4.5 \times 10^{9}$ years. The ratio of number of nuclei of $P b$ to $U^{238}$ after a time of $1.5 \times 10^{9}$ years.

(Given 21/3 = 1.26)

At radioactive equilibrium, the ratio between the number of the atoms of two radioactive elements (X) and (Y) was found to be $3.2 \times 10^{9}$ : 1 respectively. If half-life of the element (X) is $1.6 \times 10^{10}$ years, then half-life of the element (Y) would be

Q. The number of natural numbers which lie between

10^{8}

and

10^{9}

which have products of their digits as 6 is___.

Q. The solution of the following by using suitable identities:

109 \times 108

is ------

Join BYJU'S Learning Program

What is steady state number of I after a long time?

The correct option is B 3.2×109dNdt = NλGiven that: t1/2=25 min=1500 sλ=ln(2)t1/2=0.6931500N = dN/dtλN = 1.5×106×1500.693 =3.2×109

What is steady state number of $I$ after a long time?

The correct option is B $3.2 \times 10^{9}$
$\frac{d N}{d t}$ = $N λ$
Given that: $t_{1 / 2} = 25 m i n = 1500 s$
$λ = \frac{l n (2)}{t_{1 / 2}} = \frac{0.693}{1500}$
$N$ $= \frac{d N / d t}{λ}$
$N$ $= \frac{1.5 \times 10^{6} \times 1500}{.693}$ $= 3.2 \times 10^{9}$