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Alpha, beta, and gamma

Research begun by people such as New Zealander Ernest Rutherford soon after the discovery of nuclear radiation indicated that different types of rays are emitted. Eventually, three types were distinguished and named alpha α size 12{ left (α right )} {} , beta β size 12{ left (β right )} {} , and gamma γ size 12{ left (γ right )} {} , because, like x-rays, their identities were initially unknown. [link] shows what happens if the rays are passed through a magnetic field. The γ size 12{γ} {} s are unaffected, while the α size 12{γ} {} s and β size 12{β} {} s are deflected in opposite directions, indicating the α size 12{α} {} s are positive, the β size 12{β} {} s negative, and the γ size 12{γ} {} s uncharged. Rutherford used both magnetic and electric fields to show that α size 12{α} {} s have a positive charge twice the magnitude of an electron, or + 2 q e size 12{+2 lline q rSub { size 8{e} } rline } {} . In the process, he found the α size 12{γ} {} s charge to mass ratio to be several thousand times smaller than the electron’s. Later on, Rutherford collected α size 12{γ} {} s from a radioactive source and passed an electric discharge through them, obtaining the spectrum of recently discovered helium gas. Among many important discoveries made by Rutherford and his collaborators was the proof that α size 12{γ} {} radiation is the emission of a helium nucleus . Rutherford won the Nobel Prize in chemistry in 1908 for his early work. He continued to make important contributions until his death in 1934.

The figure shows north and south poles of a magnet through which three rays labeled as alpha beta and gamma are passed. After passing through a magnetic field the alpha ray is slightly deflected toward the right. The beta ray is deflected toward the left and the gamma ray is not deflected.
Alpha, beta, and gamma rays are passed through a magnetic field on the way to a phosphorescent screen. The α size 12{γ} {} s and β size 12{β} {} s bend in opposite directions, while the γ size 12{γ} {} s are unaffected, indicating a positive charge for α size 12{γ} {} s, negative for β size 12{β} {} s, and neutral for γ size 12{γ} {} s. Consistent results are obtained with electric fields. Collection of the radiation offers further confirmation from the direct measurement of excess charge.

Other researchers had already proved that β size 12{β} {} s are negative and have the same mass and same charge-to-mass ratio as the recently discovered electron. By 1902, it was recognized that β size 12{β} {} radiation is the emission of an electron . Although β size 12{β} {} s are electrons, they do not exist in the nucleus before it decays and are not ejected atomic electrons—the electron is created in the nucleus at the instant of decay.

Since γ size 12{γ} {} s remain unaffected by electric and magnetic fields, it is natural to think they might be photons. Evidence for this grew, but it was not until 1914 that this was proved by Rutherford and collaborators. By scattering γ size 12{γ} {} radiation from a crystal and observing interference, they demonstrated that γ size 12{γ} {} radiation is the emission of a high-energy photon by a nucleus . In fact, γ size 12{γ} {} radiation comes from the de-excitation of a nucleus, just as an x ray comes from the de-excitation of an atom. The names " γ size 12{γ} {} ray" and "x ray" identify the source of the radiation. At the same energy, γ size 12{γ} {} rays and x rays are otherwise identical.

Properties of nuclear radiation
Type of Radiation Range
α size 12{α} {} -Particles A sheet of paper, a few cm of air, fractions of a mm of tissue
β size 12{β} {} -Particles A thin aluminum plate, or tens of cm of tissue
γ size 12{γ} {} Rays Several cm of lead or meters of concrete

Ionization and range

Two of the most important characteristics of α size 12{α} {} , β size 12{β} {} , and γ size 12{γ} {} rays were recognized very early. All three types of nuclear radiation produce ionization in materials, but they penetrate different distances in materials—that is, they have different ranges . Let us examine why they have these characteristics and what are some of the consequences.

Questions & Answers

sound waves can be modeled as a change in pressure ,why is the change on in pressure used and not the actual pressure
Dotto Reply
what is the best
Kelly Reply
Water,air,fire
Maung
I am a university student of Myanmar.I am first year,first semester.I want to learn about physics.
Maung
two charges qA and qB are separated by a distance x. if we double the distance between the charges and triple the magnitude of the charge A, what happens to the magnitude of the force that charge A exerts on charge B. what happens to the magnitude of the force that charge B exerts on charge A
tanla Reply
how to get mcq and essay?
Owen Reply
what is force
Ibrahim Reply
force is a pull or push action on an object or a body.
joseph
what is a significant figure? and give example
Frederick
numerical chapter number 3
Sajid Reply
joined
Ibrahim
a reflected ray on a mirror makes an angle of 20degree with the incident ray when the mirror is rotated 15degree what angle will the incident ray now make with the reflected ray
Akinyemi Reply
what is simple harmonic motion
Solomon Reply
how vapour pressure of a liquid lost through convection
Yomzi Reply
Roofs are sometimes pushed off vertically during a tropical cyclone, and buildings sometimes explode outward when hit by a tornado. Use Bernoulli’s principle to explain these phenomena.
Aliraza Reply
Plz answer the question ☝️☝️
Aliraza
what's the basic si unit of acceleration
ELLOIN Reply
Explain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both.
Fabian Reply
Insulators (nonmetals) have a higher BE than metals, and it is more difficult for photons to eject electrons from insulators. Discuss how this relates to the free charges in metals that make them good conductors.
Muhammad Reply
Is the photoelectric effect a direct consequence of the wave character of EM radiation or of the particle character of EM radiation? Explain briefly.
Muhammad
Determine the total force and the absolute pressure on the bottom of a swimming pool 28.0m by 8.5m whose uniform depth is 1 .8m.
Henny Reply
how solve this problem?
Foday
P(pressure)=density ×depth×acceleration due to gravity Force =P×Area(28.0x8.5)
Fomukom
for the answer to complete, the units need specified why
muqaddas Reply
That's just how the AP grades. Otherwise, you could be talking about m/s when the answer requires m/s^2. They need to know what you are referring to.
Kyle
Practice Key Terms 8

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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