Answer to Question #210781 in Mechanics | Relativity for Hailay

Question

The β− particles emitted in the decay of 3 H (tritium) interact with matter to create light in a glow-in-the-dark exit sign. At the time of manufacture, such a sign contains 15.0 Ci of 3 H . (a) What is the mass of the tritium? (b) What is its activity 5.00 y after manufacture? 

Answer 1

(a) The emission of this sign at the start can be calculated as:

$R = \lambda N_0= \frac{0.693N_0}{t_{1/2}}=\frac{0.693}{t_{1/2}}\frac{m_0N_A}{M_{molar}}$

From there, we know that R = 15 Ci (1 Ci = 3.70 X 10¹⁰ Bq= 3.70 X 10¹⁰ decays/s), and that for ³H the molar mass M_molar=3.016 g/mol and the half-time for the β-decay t_1/2 = 12.43 years:

$R =\frac{0.693}{t_{1/2}}\frac{m_0N_A}{M_{molar}} \implies m_0=\frac{t_{1/2}\cdot R\cdot M_{molar}}{0.693N_A}$

$m_0=\cfrac{(12.43\,years(\frac{3.156\times10^{7}s}{1\,year}))(15\,Ci(\frac{3.70\times10^{10}Bq}{1\,Ci})(\frac{1\,atom/s}{1\,Bq}))(3.016\,g/mol)}{0.693(6.022\times 10^{23}\,atoms/mol)}$

$m_0=\frac{(12.43)(3.156)(15)({3.70})(3.016)}{0.693(6.022)}\times10^{7+10-23}\,\text{g of } ^3H$

$m_0=1573.47\times10^{-6}\,\text{g of }^3H=1.573\times10^{-3}\,\text{g of }^3H$

(b) After five years, we can calculate the mass by using:

$m(t)=m_0e^{-\lambda t}=m_0e^{-\cfrac{0.693}{t_{1/2}}t}$

We substitute m₀ as the start mass calculated before and t=5 years to find:

$m(\text{t=5 years})=(1.573\times10^{-3}g)e^{- \large{(\frac{5\,years}{12.43\,years})(0.693)}}$

$m(\text{t=5 years})=(1.573\,mg)e^{- \large{\frac{(5)(0.693)}{12.43}}}=(1.573\,mg)e^{- 0.278761...}$

$m(\text{t=5 years})=(1.573\,mg)(0.75672)=1.190\,\text{mg of }^3H$

In conclusion, we find that (a) the initial mass of ³H on the sign is m₀=1.573 mg and (b) after five years the mass decreases to m_{(5 years)}=1.190 mg of the ³H isotope that has a β-decay.

Reference:

• Young, H. D., & Freedman, R. A. (2015). University Physics with Modern Physics and Mastering Physics (p. 1632). Academic Imports Sweden AB.