Aproton(mass1.6726×10−27kg)andaneutron(mass1.6749×10−27kg)at rest interact to form a deuterium nucleus (heavy hydrogen).In this process,a gamma ray (high energy photon) is emitted.The energy of this gamma ray is measured to be2.2MeV≡2

Aproton(mass1.6726×10−27kg)andaneutron(mass1.6749×10−27kg)at rest interact to form a deuterium nucleus (heavy hydrogen).In this process,a gamma ray (high energy photon) is emitted.The energy of this gamma ray is measured to be2.2MeV≡2
Aproton(mass1.6726×10−27kg)andaneutron(mass1.6749×10−27kg)at rest interact to form a deuterium nucleus (heavy hydrogen).In this process,a gamma ray (high energy photon) is emitted.The energy of this gamma ray is measured to be
2.2MeV≡2.2×106 eV.
a) What is the mass of the deuteron?Assume that you can neglect the small kinetic energy of the deuteron acquires after emitting the gamma ray,but keep all five significant figures as you calculate.
b) If we include the emitted gamma ray along with the proton and neutron (now bound in the deuteron) in the final state of the system to which we apply the momentum principle,we conclude that the system’s momentum must be conserved.That is,the deuteron must recoil with momentum equal and opposite to the momentum of the emitted gamma ray.Estimate the kinetic energy of the recoiling deuteron and verify that it is indeed small compared to the energy of the gamma ray,as assumed in part a).

Aproton(mass1.6726×10−27kg)andaneutron(mass1.6749×10−27kg)at rest interact to form a deuterium nucleus (heavy hydrogen).In this process,a gamma ray (high energy photon) is emitted.The energy of this gamma ray is measured to be2.2MeV≡2
e=mc^2,也就是说在忽略重核动能的假设下,系统损失质量为3.91643*10^-30kg最后的重量为3.34358*10^-27kg.
下面是第二问,光子是微观物质,所以不能用经典力学解释,就用普朗克的公式.e=hν=hc/λ=pc.,p=1.17493*10^-21.我们在研究重核是则采用经典力学的方法.因为是估算吗.最后解出动能是4.12868*10^-16.而光子的能量是3.52479*10^-13,相差三个数量级,符合足够小的估算标准.
下面是具体演算过程
In[1]:= 2.2*10^6*1.60217653*10^-19
Out[1]= 3.52479*10^-13
In[3]:= 3.5247883659999994*10^-13/(9*10^16)
Out[3]= 3.91643*10^-30
In[4]:= 1.6726*10^-27 + 1.6749*10^-27 - 3.916431517777777*10^-30
Out[4]= 3.34358*10^-27
In[5]:= 3.5247883659999994`*^-13/(3*10^8)
Out[5]= 1.17493*10^-21
In[6]:= 1.1749294553333331`*^-21/3.3435835684822224`*^-27
Out[6]= 351398.
In[7]:= 351398^2*3.3435835684822224`*^-27
Out[7]= 4.12868*10^-16
以上单位都是国际标准单位.注意单位换算.

质子(质量是1.6726×10−27 kg) 中子(质量是1.6749×10−27公斤)相互作用形成 氘核(重型氢)。 在这个过程中,放出一个伽马(高能光子)。伽马射线的能量≡2.2 MeV =2.2×10的6次方电子伏特。
一)氘核（重氢核）质量是多少?假定你可以忽略的小动能氘核（重氢核）获得在放射出伽马射线,结果保留五位数。
b)如果我们考虑随着质子和中...

全部展开

质子(质量是1.6726×10−27 kg) 中子(质量是1.6749×10−27公斤)相互作用形成 氘核(重型氢)。 在这个过程中,放出一个伽马(高能光子)。伽马射线的能量≡2.2 MeV =2.2×10的6次方电子伏特。
一)氘核（重氢核）质量是多少?假定你可以忽略的小动能氘核（重氢核）获得在放射出伽马射线,结果保留五位数。
b)如果我们考虑随着质子和中子发出伽马射线的系统的最终状态,我们运用动量原理,得出系统的冲力必须考虑的的。那么氘核（重氢核）动量必须与伽马射线动量 大小相等、方向相反。试估算的氘核（重氢核）的动能并验证它小于伽马射线的能量的。

收起

完全看不懂