Happy Rads! (Part 2 of 2)
[Continued from this post.]
It occurred to me many times, while listening to the Geiger counter chattering away beside me, to wonder just how fast the beta particles were traveling once their parent nuclei had spat them out. And I looked all over for the answer, but nobody, it seems, was willing to part with such information. The maximum decay energies were readily available, but because these were provided in the rather cryptic units of megaelectron volts (MeV), they did little to help. One electron volt (eV) is the kinetic energy acquired by an electron as it accelerates through a potential difference of one volt. As you can imagine, an eV is a piddling small quantity of energy-so small, in fact, that about 700,000,000,000,000,000,000,000 of them, or 7 × 1023, will fit into a single Calorie (of the nutritional variety). Thus, even a million of them (1 MeV) don't add up to much. An electron, on the other hand, is exceedingly insubstantial, so a MeV or two might be sufficient to accelerate one almost to the speed of light. Or perhaps only to a few thousand meters per second. Who knows? I sure didn't.
The table below lists the maximum decay energies of my four beta emitters. I converted the units to kiloelectron volts (keV; 1000 keV = 1 MeV) so that the numbers would fall into a convenient range.
Nuclide Beta Energy (keV)
H-3 18.6
C-14 156.5
S-35 167.4
P-32 1710How to convert these energies to velocities? We could naïvely apply Newton's formula for kinetic energy: E = ½mv2, where E is kinetic energy, m is mass, and v is velocity in units of meters per second. [First, we'd need to convert the kinetic energy from eV to units of joules (J; 1 eV = 1.602 × 10-19 J), and express the mass of the electron in kilograms (9.109 × 10-31 kg).] This approach will provide the correct answer if our beta particle is moving very slowly relative to the speed of light (c)-plodding along, as it were. However, if the true velocity is close the speed of light, the answer we get might not even be close. And if the electron is really zipping along, Newton's formula might return a speed faster than c, which we know is impossible, pending a major revision of the laws of physics.
[Editor's Note: Physicist jedibl has kindly pointed out an error in my thinking in what follows. I should have formulated things in terms of relativistic energy, rather than in terms of mass. However, since I appear to have come up with the right answer, anyway, I'll leave my discussion as is until I do some reading and become conversant enough on the subject to revise it properly.]
According to Einstein's Theory of Special Relativity, as an electron or other mass-containing body is accelerated to an appreciable fraction of c, relative to the observer, its mass begins to rise appreciably. As the electron is accelerated closer and closer to c, more and more of the kinetic energy supplied to it goes into increasing the mass, and less and less into increasing the speed. Because the mass approaches infinity as the velocity approaches c, we can never quite push even an itty-bitty electron all the way to the speed of light. However, by enlisting an instrument like the Tevatron particle accelerator at Fermilab in Illinois, we can get pretty damn close. Powered by hundreds of gigantic, superconducting magnets arranged in a four-mile ring, the Tevatron can fling protons beneath the Midwestern countryside to within a few meters per second of the "fastest speed there is" (to quote Monty Python). And a proton is more than 1800 times as massive than an electron.
The relationship between speed and mass has been known for over a century, thanks again to Einstein:
__________
M = M0 / √ 1 - v2/c2
where M is the "relativistic mass" of an object, M0 is the mass of the object at rest, v is the speed of the object and c is the speed of light in a vacuum ( about 2.998 × 108 meters per second).
Now suppose we already know the ratio of the relativistic mass to the rest mass: M/M0. For simplicity, let's call this ratio Rm, the "mass ratio." (Please note: I'm making up the notation as I go. Please do not confuse my ad hoc symbols for anything officially sanctioned.) Suppose we're interested in the ratio of the speed of the particle to lightspeed; using the same notation we denote this quantity as v/c. Rearranging the equation above gives us
____________
v/c = √ 1 - (1/Rm2)
and we can easily calculate the velocity ratio, relative to c, from the mass ratio relative to the rest mass.
All that is nice and elegant, but still doesn't quite answer my question, for we still need a way to translate the decay energy into a relativistic mass for the electrons. I thought about this, off and on, for a while, consigning the problem to the back burner of my cerebral cortex; until one evening, a Profound Revelation struck like a bolt from the blue. You see, the kinetic energy added to an electron during beta decay isn't merely equivalent to mass: it literally is mass. In other words, the relativistic mass of a beta particle is equal to the rest mass of an electron, plus the mass-equivalent of the decay energy. It only remains, then, to convert the decay energy to mass by running it through E = mc2, and to find the ratio of the rest mass plus kinetic-energy "mass" to the rest mass alone. We can simplify matters even further by leaving the maximum decay energy in units of keV, and instead converting the rest mass of an electron into the same units. As it happens, we don't even need to do the conversion: the mass-energy equivalent of an electron is commonly quoted in reference books and all over the Web. According to the National Institute of Standards and Technology, the value is 510.998910 ± 0.000013 keV. Let's use 511.0 keV as an approximation-more than close enough for our purposes.
It now remains only to run our beta-decay energies through the two equations. Take, for example, tritium, with a maximum energy of 18.6 keV. Add that to the 511.0 keV electron rest mass equivalent, and divide the total mass-equivalent of 529.6 keV by the rest mass to obtain a v/c of 1.034-a gain of 3.4% in relativistic mass. Not too much, it seems; but this slight gain translates to a value of over 0.06 for 1 - (1/Rm)2, and when we take the square root, that translates to v/c = 0.254. Amazing-the weak beta particles from tritium, which can't even smash their way through your skin, are traveling at up to a quarter of the speed of light. As we move up to the next weakest beta particles, from 14C, the relativistic mass increase really takes its toll on the added speed: raising the maximum decay energy by more than eightfold, from 18.6 to 156.5 keV, increases the velocity by a factor of only 2.5, to 0.63c (see table below). And because we're already well over half the speed of light, there isn't much room to accelerate further. Raising the energy by another order of magnitude, to 1710 keV for 32P, only adds another 50% or so to the speed. Below are the values of Rm and v/c for all four beta emitters:
Nuclide Beta Energy (keV) Rm v/c
H-3 18.6 1.034 0.254
C-14 156.5 1.284 0.627
S-35 167.4 1.304 0.642
P-32 1710 4.103 0.970What if we fail to take relativity into account, and use the Newtonian formula, E = ½ mv2? In the case of 3H, the approximation is not too bad: the uncorrected velocity estimate is 0.270c, about 6 percent above the true value accounting for relativity. The discrepancy is more severe with 14C and 35S: the speeds in Newton's universe would be 0.783c and 0.809c, respectively, which overestimate the actual speeds by about 25 percent. At velocities as high as that of 32P decays, the mass correction is clearly essential, as Newton would assign this particle the speed of 2.59c-not only wildly inaccurate, but physically impossible, to boot.
You may be wondering, about now, just how fast these subatomic particles can travel when given a really powerful boost. Beta particles from 32P are the fastest radioactive decay particles with which I'm familiar. The Tevatron, on the other hand, can accelerate protons to 1 TeV (1012 eV, or 1 billion keV). Despite the proton's much greater rest mass of 938.3 MeV, that energy corresponds to a mass ratio Rm of 1067, and thus to a velocity of 0.99999956c. Forget the Plexiglas shield: any particle moving with such astonishing quickness would plow through an elephant like it was vacuum.
But for serious speed, we can look to the cosmic-ray particles. What we call "cosmic rays" are actually charged atomic nuclei and other subatomic flotsam, mostly protons, raining down on Earth from all directions with incredible energy. Most cosmic rays arrive at modest energies-say, 1014 eV or less. Nonetheless, the atmosphere over each square kilometer of the Earth's surface receives about one super-fast particle per year with an energy over 1019eV. 3 In 1992, a 3 × 1020-eV cosmic ray struck the stratosphere over Utah. The resulting shower of secondary particles-electrons, gamma rays and more exotic atomic fragments not usually observed outside high-energy colliders like the Tevatron and the new Large Hadron Collider-that reached the Earth a split second later was picked up by an instrument called the "Fly's Eye" (after its multifaceted, hemispherical structure). If that cosmic ray was a proton, it was traveling at a speed of about 0.9999999999999999999999951c-a glacier's pace shy of the speed of light-at the time it was rudely interrupted by some ill-fated air molecule.
The formula relating v/c to Rm comes in handy when thinking about whether humankind will ever travel to the stars. If we take any object at rest, and add the equivalent of its mass in kinetic energy, we will have raised its Rm to 2.0, corresponding to a velocity of 0.866c, a reasonable speed for an interstellar voyage. (The same amount is required to slow it back down to rest relative to the starting point, which would likely be necessary to explore an alien star system.)
Sounds pretty easy-but the thing to consider about E = mc2 is that c2 is a whonking huge number. Let's consider how much energy it would take to accelerate an astronaut and space suit, total 100 kg, to 0.866c Feeding 100 kg into E = mc2, we obtain an equivalent energy of 8.987 × 1018 J. Now that quantity, 8.987 × 1018 J, signifies nothing to anyone besides a physicist, so we need to find something to which to compare it that we can appreciate. For those of us who grew up during the Cold War, a nuclear explosion comes immediately to mind. The explosive yield of modern-day nuclear weapons is most conveniently expressed in units of kilotons, or one thousand tonnes of TNT. Little Boy exploded over Hiroshima with a force of between 12 and 15 kilotons, eventually killing 200,000 people including the victims of long-term effects of radiation exposure. Today's thermonuclear (hydrogen) weapons fall mostly between seven and thirty times that much power; Little Boy would now be considered a "tactical" bomb, useful only to achieve a minor, specific military goal. At the other end of the scale, the largest American hydrogen bomb test, Castle Bravo, had a yield of 15,000 kilotons, or 15 megatons. The explosion created a mushroom cloud 25 miles (40 km) high and blanketed thousands of square miles of the Pacific Ocean, and several atolls of the Marshall Island chain, with lethal radioactive fallout. It was the greatest ecological disaster in the history of US nuclear testing.
Detonation of one thousand tonnes of TNT releases a mere 4.19 × 1012 J. Consequently, 100 kg of mass, or 8.987 × 1018 J, corresponds to 2,146,000 kilotons, or 2146 megatons. To wit: the energy of 143 Castle Bravos would need to be precisely directed with 100% efficiency to send our astronaut spaceward at 87 percent of the speed of light. And that doesn't include supplies, crewmates, the spaceship, fuel, and such. Far more energy would be necessary to get that lot going at an appreciable fraction of c.
Obviously, we have some serious challenges ahead of us to overcome before we can expect to travel to the stars.
____________________
3Clay, R. and Dawson, B. (1997). Cosmic Bullets. Cambridge, MA: Perseus Publishing.
It occurred to me many times, while listening to the Geiger counter chattering away beside me, to wonder just how fast the beta particles were traveling once their parent nuclei had spat them out. And I looked all over for the answer, but nobody, it seems, was willing to part with such information. The maximum decay energies were readily available, but because these were provided in the rather cryptic units of megaelectron volts (MeV), they did little to help. One electron volt (eV) is the kinetic energy acquired by an electron as it accelerates through a potential difference of one volt. As you can imagine, an eV is a piddling small quantity of energy-so small, in fact, that about 700,000,000,000,000,000,000,000 of them, or 7 × 1023, will fit into a single Calorie (of the nutritional variety). Thus, even a million of them (1 MeV) don't add up to much. An electron, on the other hand, is exceedingly insubstantial, so a MeV or two might be sufficient to accelerate one almost to the speed of light. Or perhaps only to a few thousand meters per second. Who knows? I sure didn't.
The table below lists the maximum decay energies of my four beta emitters. I converted the units to kiloelectron volts (keV; 1000 keV = 1 MeV) so that the numbers would fall into a convenient range.
Nuclide Beta Energy (keV)
H-3 18.6
C-14 156.5
S-35 167.4
P-32 1710How to convert these energies to velocities? We could naïvely apply Newton's formula for kinetic energy: E = ½mv2, where E is kinetic energy, m is mass, and v is velocity in units of meters per second. [First, we'd need to convert the kinetic energy from eV to units of joules (J; 1 eV = 1.602 × 10-19 J), and express the mass of the electron in kilograms (9.109 × 10-31 kg).] This approach will provide the correct answer if our beta particle is moving very slowly relative to the speed of light (c)-plodding along, as it were. However, if the true velocity is close the speed of light, the answer we get might not even be close. And if the electron is really zipping along, Newton's formula might return a speed faster than c, which we know is impossible, pending a major revision of the laws of physics.
[Editor's Note: Physicist jedibl has kindly pointed out an error in my thinking in what follows. I should have formulated things in terms of relativistic energy, rather than in terms of mass. However, since I appear to have come up with the right answer, anyway, I'll leave my discussion as is until I do some reading and become conversant enough on the subject to revise it properly.]
According to Einstein's Theory of Special Relativity, as an electron or other mass-containing body is accelerated to an appreciable fraction of c, relative to the observer, its mass begins to rise appreciably. As the electron is accelerated closer and closer to c, more and more of the kinetic energy supplied to it goes into increasing the mass, and less and less into increasing the speed. Because the mass approaches infinity as the velocity approaches c, we can never quite push even an itty-bitty electron all the way to the speed of light. However, by enlisting an instrument like the Tevatron particle accelerator at Fermilab in Illinois, we can get pretty damn close. Powered by hundreds of gigantic, superconducting magnets arranged in a four-mile ring, the Tevatron can fling protons beneath the Midwestern countryside to within a few meters per second of the "fastest speed there is" (to quote Monty Python). And a proton is more than 1800 times as massive than an electron.
The relationship between speed and mass has been known for over a century, thanks again to Einstein:
__________
M = M0 / √ 1 - v2/c2
where M is the "relativistic mass" of an object, M0 is the mass of the object at rest, v is the speed of the object and c is the speed of light in a vacuum ( about 2.998 × 108 meters per second).
Now suppose we already know the ratio of the relativistic mass to the rest mass: M/M0. For simplicity, let's call this ratio Rm, the "mass ratio." (Please note: I'm making up the notation as I go. Please do not confuse my ad hoc symbols for anything officially sanctioned.) Suppose we're interested in the ratio of the speed of the particle to lightspeed; using the same notation we denote this quantity as v/c. Rearranging the equation above gives us
____________
v/c = √ 1 - (1/Rm2)
and we can easily calculate the velocity ratio, relative to c, from the mass ratio relative to the rest mass.
All that is nice and elegant, but still doesn't quite answer my question, for we still need a way to translate the decay energy into a relativistic mass for the electrons. I thought about this, off and on, for a while, consigning the problem to the back burner of my cerebral cortex; until one evening, a Profound Revelation struck like a bolt from the blue. You see, the kinetic energy added to an electron during beta decay isn't merely equivalent to mass: it literally is mass. In other words, the relativistic mass of a beta particle is equal to the rest mass of an electron, plus the mass-equivalent of the decay energy. It only remains, then, to convert the decay energy to mass by running it through E = mc2, and to find the ratio of the rest mass plus kinetic-energy "mass" to the rest mass alone. We can simplify matters even further by leaving the maximum decay energy in units of keV, and instead converting the rest mass of an electron into the same units. As it happens, we don't even need to do the conversion: the mass-energy equivalent of an electron is commonly quoted in reference books and all over the Web. According to the National Institute of Standards and Technology, the value is 510.998910 ± 0.000013 keV. Let's use 511.0 keV as an approximation-more than close enough for our purposes.
It now remains only to run our beta-decay energies through the two equations. Take, for example, tritium, with a maximum energy of 18.6 keV. Add that to the 511.0 keV electron rest mass equivalent, and divide the total mass-equivalent of 529.6 keV by the rest mass to obtain a v/c of 1.034-a gain of 3.4% in relativistic mass. Not too much, it seems; but this slight gain translates to a value of over 0.06 for 1 - (1/Rm)2, and when we take the square root, that translates to v/c = 0.254. Amazing-the weak beta particles from tritium, which can't even smash their way through your skin, are traveling at up to a quarter of the speed of light. As we move up to the next weakest beta particles, from 14C, the relativistic mass increase really takes its toll on the added speed: raising the maximum decay energy by more than eightfold, from 18.6 to 156.5 keV, increases the velocity by a factor of only 2.5, to 0.63c (see table below). And because we're already well over half the speed of light, there isn't much room to accelerate further. Raising the energy by another order of magnitude, to 1710 keV for 32P, only adds another 50% or so to the speed. Below are the values of Rm and v/c for all four beta emitters:
Nuclide Beta Energy (keV) Rm v/c
H-3 18.6 1.034 0.254
C-14 156.5 1.284 0.627
S-35 167.4 1.304 0.642
P-32 1710 4.103 0.970What if we fail to take relativity into account, and use the Newtonian formula, E = ½ mv2? In the case of 3H, the approximation is not too bad: the uncorrected velocity estimate is 0.270c, about 6 percent above the true value accounting for relativity. The discrepancy is more severe with 14C and 35S: the speeds in Newton's universe would be 0.783c and 0.809c, respectively, which overestimate the actual speeds by about 25 percent. At velocities as high as that of 32P decays, the mass correction is clearly essential, as Newton would assign this particle the speed of 2.59c-not only wildly inaccurate, but physically impossible, to boot.
You may be wondering, about now, just how fast these subatomic particles can travel when given a really powerful boost. Beta particles from 32P are the fastest radioactive decay particles with which I'm familiar. The Tevatron, on the other hand, can accelerate protons to 1 TeV (1012 eV, or 1 billion keV). Despite the proton's much greater rest mass of 938.3 MeV, that energy corresponds to a mass ratio Rm of 1067, and thus to a velocity of 0.99999956c. Forget the Plexiglas shield: any particle moving with such astonishing quickness would plow through an elephant like it was vacuum.
But for serious speed, we can look to the cosmic-ray particles. What we call "cosmic rays" are actually charged atomic nuclei and other subatomic flotsam, mostly protons, raining down on Earth from all directions with incredible energy. Most cosmic rays arrive at modest energies-say, 1014 eV or less. Nonetheless, the atmosphere over each square kilometer of the Earth's surface receives about one super-fast particle per year with an energy over 1019eV. 3 In 1992, a 3 × 1020-eV cosmic ray struck the stratosphere over Utah. The resulting shower of secondary particles-electrons, gamma rays and more exotic atomic fragments not usually observed outside high-energy colliders like the Tevatron and the new Large Hadron Collider-that reached the Earth a split second later was picked up by an instrument called the "Fly's Eye" (after its multifaceted, hemispherical structure). If that cosmic ray was a proton, it was traveling at a speed of about 0.9999999999999999999999951c-a glacier's pace shy of the speed of light-at the time it was rudely interrupted by some ill-fated air molecule.
The formula relating v/c to Rm comes in handy when thinking about whether humankind will ever travel to the stars. If we take any object at rest, and add the equivalent of its mass in kinetic energy, we will have raised its Rm to 2.0, corresponding to a velocity of 0.866c, a reasonable speed for an interstellar voyage. (The same amount is required to slow it back down to rest relative to the starting point, which would likely be necessary to explore an alien star system.)
Sounds pretty easy-but the thing to consider about E = mc2 is that c2 is a whonking huge number. Let's consider how much energy it would take to accelerate an astronaut and space suit, total 100 kg, to 0.866c Feeding 100 kg into E = mc2, we obtain an equivalent energy of 8.987 × 1018 J. Now that quantity, 8.987 × 1018 J, signifies nothing to anyone besides a physicist, so we need to find something to which to compare it that we can appreciate. For those of us who grew up during the Cold War, a nuclear explosion comes immediately to mind. The explosive yield of modern-day nuclear weapons is most conveniently expressed in units of kilotons, or one thousand tonnes of TNT. Little Boy exploded over Hiroshima with a force of between 12 and 15 kilotons, eventually killing 200,000 people including the victims of long-term effects of radiation exposure. Today's thermonuclear (hydrogen) weapons fall mostly between seven and thirty times that much power; Little Boy would now be considered a "tactical" bomb, useful only to achieve a minor, specific military goal. At the other end of the scale, the largest American hydrogen bomb test, Castle Bravo, had a yield of 15,000 kilotons, or 15 megatons. The explosion created a mushroom cloud 25 miles (40 km) high and blanketed thousands of square miles of the Pacific Ocean, and several atolls of the Marshall Island chain, with lethal radioactive fallout. It was the greatest ecological disaster in the history of US nuclear testing.
Detonation of one thousand tonnes of TNT releases a mere 4.19 × 1012 J. Consequently, 100 kg of mass, or 8.987 × 1018 J, corresponds to 2,146,000 kilotons, or 2146 megatons. To wit: the energy of 143 Castle Bravos would need to be precisely directed with 100% efficiency to send our astronaut spaceward at 87 percent of the speed of light. And that doesn't include supplies, crewmates, the spaceship, fuel, and such. Far more energy would be necessary to get that lot going at an appreciable fraction of c.
Obviously, we have some serious challenges ahead of us to overcome before we can expect to travel to the stars.
____________________
3Clay, R. and Dawson, B. (1997). Cosmic Bullets. Cambridge, MA: Perseus Publishing.