推荐回答(4个)
在大质量恒星演化到生命的末期,当低原子序数的元素燃烧完时就会发生超新星爆发。那么地球上有那么多的铁,地球会不会很危险?其实大可不必担心,地球上的铁是不会引起超新星爆发的,至于为什么?我们得先从恒星演化说起。
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前奏——一个太阳质量恒星的诞生
当原始气体云在引力作用下发生坍缩,在收缩过程中核心温度开始上升,当气体云坍缩到一定程度,核心温度达到10000K左右,氢氦等原子开始电离,当气体云进一步坍缩,核心温度进一步升高,达到约10000000K(1000万开尔文),核心电离出的氢核会发生核聚变反应结合成氦核并释放出大量辐射能量。
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辐射出的能量产生高温热膨胀,阻止了气体云继续坍缩,核心温度和压力会达到一定不断平衡,核心的氢开始稳定地燃烧,中心辐射开始往外传播,使外层气体温度上升,当外层达到热平衡状态,开始以热辐射形式向外辐射,一颗发光发热的恒星正式诞生。
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主序星阶段——恒星核聚变
对于一个太阳质量的恒星,当恒星燃烧完氢,由于引力坍缩产生的压力和温度未能直接点燃核心的氦,氦核会在核心堆积并在极大的压力下进入电子简并态,核心急剧收缩温度也急剧上升,在电子简并态的巨大压力和温度下部分氦终于被点燃,核心由于处于电子简并态,无法产生热膨胀,导致核心温度失控地升高使更多的氦剧烈燃烧,这就产生了氦闪,每错,那就是科幻电影《流浪地球》里地球要逃离太阳的原因。由于氦闪过于剧烈,会导致大量完成气体被从太阳表面抛出,同样会进入红巨星阶段,而在高温下,恒星核心终于脱离电子简并态并继续燃烧,直到氦全部烧完,恒星会再次进入以碳和氧为主的电子简并态,这个状态称为白矮星。我们熟悉的天狼星的伴星——天狼星B就是一颗白矮星。
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对于一颗太阳质量的恒星,它是无法燃烧到核心产生铁的阶段的,但对于一颗大质量恒星,比如超过25倍太阳质量的恒星,由于引力坍缩产生的巨大压力,核心氦不会发生氦闪,而是在氢燃烧结束后,在核心收缩过程中就产生足够的温度点燃核心的氦。在燃烧完氦后也会直接过渡到碳、氖、氧、硅等元素的核聚变……
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大质量恒星末期——超新星爆发
对于一颗质量大于10倍太阳质量演化到末期的恒星,其中心的铁核开始堆积,核心会原来越重,导致更激烈的燃烧,但当低序数的原子核核聚变结束,无法抵御坍缩,而核心铁核由于核聚变所需的温度超过了其核聚变后所能释放的温度,因此无法通过核聚变释放能量来抵御坍缩,恒星最终会向内坍缩,外层气体会以近乎自由落体向内跌落,最终撞在处于电子简并态的铁核上并被反弹出去,这类似于弹性碰撞,这些外层物质跌落有多快被弹飞就有躲开,最终恒星一一次超新星爆发炸开。
![](/picurl?url=https%3A%2F%2Fiknow-pic.cdn.bcebos.com%2Ff603918fa0ec08fa7044d9da49ee3d6d55fbda03%3Fx-bce-process%3Dimage%252Fresize%252Cm_lfit%252Cw_600%252Ch_800%252Climit_1%252Fquality%252Cq_85%252Fformat%252Cf_auto)
当剩余质量足够大,核心简并态铁核会坍缩到中子简并态甚至黑洞。
结语——让人放心的答案
你会发现恒星之所以最终发生超新星爆发,并不是因为铁爆炸了,而是由于铁不能爆炸,不能通过核聚变释放能量抵御坍缩
大质量的恒星最后在聚变形成大量的铁元素之后,会形成一种元素分层分布的洋葱结构,中心为铁核,向外依次为硅层、镁层、氧层、碳层、氦层、氢层。当聚变形成的铁核质量超过钱德拉塞卡极限(1.44倍太阳质量)时,就会引发铁核的坍缩,重力势能被释放,部分铁原子重新被离解为氦原子,外层电子在巨大压力下被压进氦核,开启恒星中子化的进程,同时释放大量中微子。而在此过程发生之后,继续进行恒星的演化,即由坍缩引发的爆炸。当坍缩的外层物质在向内核移动的过程中,遇到处在向中子化方向发展的内核时,就会产生超能量的激发波,激发波向外层进行反弹,将恒星外层的物质瞬间剥离,从而出现光度迅速上升、恒星物质大量散发的爆炸现象,也就是我们常说的超新星爆炸。
因为铁元素在极高的温度和极高的密度下,与自由中子、自由电子、质子及其它原子核发生反应,能够产生出92号元素铀之前的所有元素,会引起超新星爆发
铁元素是核聚变的尽头,再聚变就不能释放能量了,要吸收能量了。所以不能提供热量供恒星继续运行,恒星就会急剧坍缩,坍缩过程中产生大量热量,又会导致恒星急剧膨胀,亮度骤增,也就是超新星爆发。
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