Translation examples
Kun tämä massavaje tiedetään, voidaan Einsteinin kaavaa (E = mc²) käyttäen laskea minkä tahansa ytimen sidosenergia.
Once this mass difference, called the mass defect or mass deficiency, is known, Einstein's mass-energy equivalence formula E = mc² can be used to compute the binding energy of any nucleus.
Kun neutronit ja protonit täyttävät tietyn kuoren atomin ytimessä, nukleonien sidosenergia saavuttaa miniminsä, ja ydin on stabiilimpi eikä hajoa niin nopeasti kuin isotoopit, joilla näitä kuoria ei ole täytetty.
So when the number of neutrons and protons completely fills the energy levels of a given shell in the nucleus, the binding energy per nucleon will reach a local maximum and thus that particular configuration will have a longer lifetime than nearby isotopes that do not possess filled shells.
Semiempiirisen massakaavan mukaan sidosenergia voidaan esittää muodossa E B = a V A − a S A 2 / 3 − a C Z ( Z − 1 ) A 1 / 3 − a A ( A − 2 Z ) 2 A + δ ( A , Z ) . {\displaystyle E_{B}=a_{V}A-a_{S}A^{2/3}-a_{C}{\frac {Z(Z-1)}{A^{1/3}}}-a_{A}{\frac {(A-2Z)^{2}}{A}}+\delta (A,Z).} Kullakin kaavan termillä on teoreettinen perusta.
The semi-empirical mass formula states that the binding energy will take the following form: E B = a V A − a S A 2 / 3 − a C Z ( Z − 1 ) A 1 / 3 − a A ( A − 2 Z ) 2 A ± δ ( A , Z ) {\displaystyle E_{B}=a_{V}A-a_{S}A^{2/3}-a_{C}{\frac {Z(Z-1)}{A^{1/3}}}-a_{A}{\frac {(A-2Z)^{2}}{A}}\pm \delta (A,Z)} Each of the terms in this formula has a theoretical basis, as will be explained below.
Olkoon A nukleonien kokonaismäärä, Z protonien lukumäärä ja N neutronien lukumäärä siten, että A = Z + N. Atomiytimen massa saadaan kaavasta m = Z m p + N m n − E B c 2 , {\displaystyle m=Zm_{p}+Nm_{n}-{\frac {E_{B}}{c^{2}}},} missä m p {\displaystyle m_{p}} ja m n {\displaystyle m_{n}} ovat protonin ja neutronin lepomassat ja E B {\displaystyle E_{B}} on ytimen sidosenergia.
In the following formula, let A be the total number of nucleons, Z the number of protons, and N the number of neutrons, so that A = Z + N. The mass of an atomic nucleus is given by m = Z m p + N m n − E B c 2 {\displaystyle m=Zm_{p}+Nm_{n}-{\frac {E_{B}}{c^{2}}}} where m p {\displaystyle m_{p}} and m n {\displaystyle m_{n}} are the rest mass of a proton and a neutron, respectively, and E B {\displaystyle E_{B}} is the binding energy of the nucleus.
Atomiytimen sähköinen sidosenergia, heikon vuorovaikutuksen energia ytimen sisällä ja atomin elektronien liike-energia vaikuttavat kaikki atomin painavaan massaan, minkä Eötvösin koe on osoittanut hyvin suurella tarkkuudella.[50
The electrostatic binding energy of the nucleus, the energy of weak interactions in the nucleus, and the kinetic energy of electrons in atoms, all contribute to the gravitational mass of an atom, as has been confirmed to high precision in Eötvös type experiments.[50
Tätä raskaammilla alkuaineilla sidosenergia vähenee ydinhiukkasta kohti atomimassan kasvaessa.
Finally, in elements heavier than xenon, there is a decrease in binding energy per nucleon as atomic number increases.
Ytimen sidosenergia voidaan laskea ytimen massan ja sen muodostavien protoneiden ja neutroneiden massojen erotuksena.
Nuclear binding energy can be computed from the difference in mass of a nucleus, and the sum of the masses of the number of free neutrons and protons that make up the nucleus.
Semiempiirinen massakaava antaa hyvät arviot atomimassoille, mutta se ei selitä, miksi tietyillä maagisilla luvuilla sidosenergia nukleonia kohti on korkeampi ja atomiydin stabiilimpi kuin massakaava ennustaa.
The SEMF gives a good approximation for atomic masses and several other effects, but does not explain the appearance of magic numbers of protons and neutrons, and the extra binding-energy and measure of stability that are associated with these numbers of nucleons.
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