The glow-in-the-dark dials on some watches and some keychain lights shine with e

Question

The glow-in-the-dark dials on some watches and some keychain lights shine with energy provided by the decay of radioactive tritium. Tritium is a radioactive isotope of hydrogen with a half-life of 12 years. Each decay emits an electron with an energy of 19 keV. A typical new watch has tritium with a total activity of 19 MBq.

What is the speed of the emitted electron? (This speed is high enough that you’ll need to do a relativistic calculation.)

What is the power, in watts, provided by the radioactive decay process?

What will be the activity of the tritium in a watch after 4 years, assuming none escapes?

Please help me to understand how to solve this problem!!

Explanation / Answer

(a) If the energy of an electron is 13 keV, and the mass of an electron is 512 keV/c^2, then clearly this energy doesn't include the rest mass, so the energy is simply kinetic. Therefore,

K + mc^2 = E = mc^2, where = 1/(1 - (v/c)^2)

So, K + mc^2 = 19 keV + 512 keV = 531 keV.

If the electron has a mass of 512 keV/c^2, then the ratio of the total energy to rest energy is

531 keV/512 keV = 1.037

So why did I do this? Because clearly the total energy is 1.037 * rest energy = 1.037mc^2, and as I said before, E = mc^2, so

= 1.037,

1 - (v/c)^2 = (1/1.037)^2

(v/c)^2 = 1 - 0.9929 = 0.070

So, v = (0.07)^2c = 0.0049c = 1.47 * 10^6 m/s

(b) So, if 1 electron carries an energy of 19 keV, and the activity is 19*10^6 Bq (ie 19*10^6 electrons produced every second), then the power is

19*10^6 Bq * 19*10^3 eV * 1.602*10^-19 J/eV = 5.78 * 10^-8 W

(c) If the halflife is 12 years, then = ln(2)/t1/2 = 0.693/(12 years) = 0.578 1/year

The activity is defined as

A(t) = A(0)e^(-t) = (19 MBq)e^(-0.578 * 4) = 1.88 MBq

Hope this helps. Good luck.

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