The half-life of Ra^{226} is 1620 years. Calc | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The number of atoms decaying per second is given by the activity $A = \lambda N$.First,calculate the decay constant $\lambda = \frac{0.693}{T_{1/2

Vedclass · Accepted Answer

$3.61 	imes 10^{10}$

Vedclass · Answer

(C) The number of atoms decaying per second is given by the activity $A = \lambda N$.First,calculate the decay constant $\lambda = \frac{0.693}{T_{1/2}}$.Half-life $T_{1/2} = 1620 	ext{ years} = 1620 	imes 365 	imes 24 	imes 3600 	ext{ seconds} \approx 5.11 	imes 10^{10} 	ext{ s}$.The number of atoms $N$ in $1 \ g$ of $Ra^{226}$ is $N = \frac{	ext{mass}}{	ext{molar mass}} 	imes N_A = \frac{1}{226} 	imes 6.023 	imes 10^{23} \approx 2.665 	imes 10^{21} 	ext{ atoms}$.Now,calculate activity $A = \lambda N = \left( \frac{0.693}{5.11 	imes 10^{10}} ight) 	imes 2.665 	imes 10^{21} \approx 3.61 	imes 10^{10} 	ext{ decays/s}$.

The half-life of $Ra^{226}$ is $1620$ years. Calculate the number of atoms that decay in one second in $1 \ g$ of radium (Avogadro number $= 6.023 \times 10^{23}$).

Similar Questions

If the vector $a = 2\hat{i} + 3\hat{j} + 6\hat{k}$ and $b$ are collinear and $|b| = 21$,then $b$ is equal to:

The period of $f(x)=\cos \left(\frac{x}{3}\right)+\sin \left(\frac{x}{2}\right)$ is (in $\pi$)

The lines represented by the equation $3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0$ and the line $4x - 3y - 2 = 0$:

If $u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)$ then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is equal to:

The transformed equation of $3x^2 + 3y^2 + 2xy = 2$,when the coordinate axes are rotated through an angle of $45^{\circ}$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module