Chapter 42: Q. 51 (page 1237)

The radium isotope ${}^{223}R a$ , an alpha emitter, has a half-life of localid="1650483334573" $11.43 d a y s$ . You happen to have a localid="1650483350108" $1.0 g$ cube of ${}^{223}R a$ , so you decide to use it to boil water for tea. You fill a well-insulated container with $100 m L$ of water at $18 ° C$ and drop in the cube of radium. How long will it take the water to boil?

Short Answer

Expert verified

Time taken to boil water is, $∆ t = 18.9 s$

Step by step solution

Given information

We are given that,

Half life is, $t_{\frac{1}{2}} = 11.43 d a y s = 9.876 \times 10^{5} s$

And $m = 100 m L = 0.10 L$

And ${}^{223}R a \to {}^{219}R n + {}^{4}H e$

Explanation

Energy is,

$E = [m ({}^{223}R a) - m ({}^{219}R n) - m ({}^{4}H e)] \times 931.49 M e V / u = [223.018499 u - 219.009477 u - 4.002602] \times 931.49 M e V / u = 5.98 M e V$

Energy needed to raise temperature of water at $18 ° C t o 100 ° C$

localid="1650483848390" $Q = m c ∆ t = (0.10) (\frac{1 k g}{1 L}) (4190 J / k g K) (100 K - 18 K) = 34400 J$

Now, value of $α - p a r t i c l e$ is,

localid="1650483872050" $5.98 \times 10^{6} e V \times (1.6 \times 10^{- 19} J / e V) = 9.57 \times 10^{- 13} J$

No. of decays,

localid="1650483894424" $N = \frac{Q}{α - p a r t i c l e} = \frac{34400 J}{9.57 \times 10^{- 13} J} = 3.59 \times 10^{16}$

Simplify

No. of radium atoms is,

$N_{0} = \frac{1 g}{223 g / m o l} \times 6.02 \times 10^{23} a t o m s / m o l = 2.7 \times 10^{21} a t o m s$

Now, $∆ t$ is given as,

localid="1650483972186" $\frac{d N}{d t} = \frac{∆ N}{∆ t} = - r N_{0} ∆ t = - \frac{∆ N}{r N_{0}} = - τ \frac{∆ N}{N_{0}} = \frac{t_{\frac{1}{2}}}{\ln (2)} \times \frac{∆ N}{N_{0}} = \frac{9.876 \times 10^{5} s}{\ln (2)} \times \frac{(- 3.59 \times 10^{16})}{2.7 \times 10^{21}} = 18.9 s$

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