You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Law of Radioactivity

If the activi...

Question

If the activity of $^{108}$ $A g$ is $3$ micro curie, the number of atoms present in it are ( $λ =$ 0.005 sec $^{- 1}$ )

2.2 x 10

^{7}

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

2.2 x 10

^{6}

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

2.2 x 10

^{5}

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

2.2 x 10

^{4}

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

Open in App

Solution

The correct option is D 2.2 x 10 $^{7}$
$λ = 0.005 s e c^{- 1}$
$1$ Micro curie $= 3.7 \times 10^{4} d p s$
Assuming all $A g$ atoms to be radioactive, and each showing
Let no of atoms present $= 1 d p s n o$
Given:
$(\frac{d N}{d t}) a c t i v i t y = λ N_{0} \Rightarrow 3.7 \times 3 \times 10^{4} d p s = 0.005 \times N_{0}$
$\Rightarrow N_{0} = 2.2 \times 10^{7} a t o m s$

Suggest Corrections

Similar questions

An engine can pump 30,000 litres of water to a vertical height of 45 metres in 10 minutes. Calculate the power. Take g= 9.8 m s^-1 [Density of water = 10³ kg m^-3 1000 liters = 1 m³]

Q. The pressure that has to be applied to the ends of a steel wire of length

10 cm

to keep its length constant when its temperature is raised by

100^{\circ} C

is:-
Given:

Y = 2 \times 10^{11} {N/m}^{2}

and

α = 1.1 \times 10^{- 5} /^{\circ} C

Q. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by

100^{\circ} C i s (F o r s t e e l Y = 2 \times 10^{11} N m^{- 2}

and

α = 1.1 \times 10^{- 5} K^{- 1})

Q. If we have a pure

N a_{2} S O_{4} .5 H_{2} O

sample and it contains

2.2 \times 10^{10}

electrons, the mass of the sample is

x \times 10^{- 14} g

. x is

Q. Equilibrium constant

(K_{C})

2 H I_{(g)} ⇌ H_{2 (g)} + I_{2 (g)}

2 \times 10^{- 4}

. What is the equilibrium concentration of

H I

, if concentration of

H_{2 (g)}

and

I_{2 (g)}

respectively are

2.2 \times 10^{- 2} M

and

2.2 \times 10^{- 4} M

Join BYJU'S Learning Program

If the activity of 108Ag is 3 micro curie, the number of atoms present in it are ( λ= 0.005 sec−1)

The correct option is D 2.2 x 107λ=0.005sec−11 Micro curie =3.7×104dpsAssuming all Ag atoms to be radioactive, and each showing Let no of atoms present =1 dps noGiven:(dNdt)activity=λN0⇒3.7×3×104dps=0.005×N0 ⇒N0=2.2×107atoms

If the activity of $^{108}$ $A g$ is $3$ micro curie, the number of atoms present in it are ( $λ =$ 0.005 sec $^{- 1}$ )