You are watching: Is mass conserved in a nuclear reaction
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edited Nov 20 "18 at 19:09
inquiry Sep 18 "13 in ~ 15:20
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$\begingroup$ This regulation apilicable just chemical reactions. If gases are commodities ,reactants shuld be take in close up door system. $\endgroup$
Mar 7 "14 in ~ 11:38
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Nuclear reactions show up to hurt both the legislations of preservation of Mass and also Energy because mass is converted into energy or angry versa.
However, the concept of mass-energy equivalence that emerges as a repercussion of Einstein"s general Theory of Relativity renders the regulations of preservation of Mass and also Energy one-of-a-kind limiting instances of the legislation of preservation of Mass-Energy. Since mass is now a form of energy, we can convert ago and soon as lengthy as there is no network loss between the two.
The equivalence of mass and also energy comes from Einstein"s well known $E=mc^2$, wherein $c$ is the speed of light (in a vacuum).
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answer Sep 18 "13 at 18:54
Ben NorrisBen Norris
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Since this thread seems to have been bumped by
Community a better answer is in order. Physicists don"t typically make a distinction between rest mass $m_0$ and moving mass $m$ but rather consider mass to it is in the invariant that the energy-momentum $4$-vector: $mc^2=\sqrtE^2-\boldsymbolp^2c^2$ wherein $m$ is the invariant massive of the system, $E$ its total energy and $\boldsymbol p$ its total momentum. $\boldsymbol p=\sum\boldsymbol p_i$, the vector sum of the momenta the the system"s constituents. Because that a massless particle, $\boldsymbol p_i=E_i\hat\boldsymbol T_i/c$, whereby $\hat\boldsymbol T_i$ is the unit vector in the direction that propagation of the particle, while for a particle with mass $m_i$ its momentum is $\boldsymbol p_i=\gamma_im_i\boldsymbol v_i$ whereby $\boldsymbol v_i$ is the particle"s velocity and $\gamma_i=1/\sqrt1-\boldsymbol v_i^2/c^2$ is that is Lorentz factor.
The massive is invariant under Lorentz revolutions which is come say that"s it"s the exact same in every inertial frames, so also when moving we think about the fixed of the device to it is in the same. As severe example that a reaction that seems to transform mass right into energy, consider $e^+e^-$ annihilation. In the rest frame of the $e^+e^-$ pair, $\boldsymbol p_+=\boldsymbol p_-=\boldsymbol0$ therefore $E_+=E_-=m_ec^2\approx0.511MeV$. So the complete energy is $E=E_++E_-=2m_ec^2$ and the complete momentum is $\boldsymbol p=\boldsymbol p_++\boldsymbol p_-=\boldsymbol0$ and so the invariant fixed of the system works the end to $m=2m_e$.
After annihilation, we obtain a pair the photons and also from momentum conservation we get that $\boldsymbol p_1=-\boldsymbol p_2$ and energy conservation tells us that $E_1=|\boldsymbol p_1|c=E_2=|\boldsymbol p_2|c=m_ec^2$ yet what occurred to the mass? Doesn"t a photon have zero mass? A photon walk in fact have zero mass, however a mechanism of 2 photons has nonzero mass uneven the photons are moving in the exact same direction. Take the $x$-direction to be the direction of take trip of the first photon and let the various other photon travel in the $xy$-plane through non-negative $y$-velocity. Then$$\beginalign\boldsymbol p^2&=p_x^2+p_y^2=(p_1x+p_2x)^2+p_2y^2=\left(\fracE_1c+\fracE_2c\cos\theta\right)^2+\left(\fracE_2c\sin\theta\right)^2\\&=\fracE_1^2c^2+\fracE_2^2c^2+2\fracE_1E_2c^2\cos\theta\endalign$$where $\theta$ is the angle in between the velocities the the to photons. So for the system of two photons,$$m^2c^4=E^2-\boldsymbol p^2c^2=\left(E_1+E_2\right)^2-E_1^2-E_2^2-2E_1E_2\cos\theta=4E_1E_2\sin^2\frac\theta2$$So the mass of the device of two photons is$$m=2\frac\sqrtE_1E_2c^2\sin\frac\theta2$$In our example, $E_1=E_2=m_ec^2$ and $\theta=180°$ for this reason we have actually $m=2m_e$, therefore the fixed didn"t readjust nor walk the power in the reaction the $e^+e^-$ annihilation. This is emphasized the the Wikipedia ar cited earlier: mass and energy don"t get interconverted yet rather are independently conserved.
If this seems choose sort the a theoretical an interpretation of mass, where perform you think most of the fixed in her body is concentrated? ns have had actually students who, after ~ a life time of body shaming answer, "In my butt?", however the ideal answer is the course: "In your atomic nuclei!" and the quarks that make up your atomic nuclei are relocating at a healthy portion of the speed of light, for this reason if us didn"t usage the relativistic $4$-vector an interpretation of mass to compute her mass us wouldn"t come up v a number that was all that close to your mass together measured by a balance or her resistance come acceleration.
After atom nuclei or molecule react, the massive of the reaction is the same to the mass of the products, and the same holds for your momentum and also energy. A well-insulated reaction courage would have the same mass ~ reaction as before, it"s only later on as warmth of reaction leaks out, carrying through it power and likewise mass that the fixed changes, immeasurably because that a chemistry reaction yet significantly for a atom reaction. The mass might be virtually unrecoverable if it"s carried away by neutrinos in a beta decay or through gravitational tide in collisions that compact objects, however it"s still the end there in the world someplace.
The masses of the individual components of the system in general won"t it is in the same for reactants together they were because that products, but masses aren"t what add, it"s the contents of the energy-momentum $4$-vector that add like $4$-vectors and from the $4$-vector we have the right to determine the massive of the system.