原子结构
国际文凭组织 · IB Diploma · 物理 · SL 知识点 16 4:14 英语旁白 · 英中内嵌字幕
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An excited gas produces separate coloured lines rather than every wavelength.
受激气体产生分离的彩色谱线，而非连续波长。
The pattern is evidence for discrete atomic energy differences.
该图样是原子能级差离散的证据。
Atoms have a small positive nucleus and electrons.
原子具有一个微小的带正电的原子核和电子。
Emission and absorption spectra arise from transitions between discrete energy levels, which are also known as energy levels.
发射光谱和吸收光谱源于离散能级之间的跃迁。
A photon energy equals the level difference and obeys E equals h f.
光子能量等于能级差，并遵循公式 E = hf。
Emission spectrum is the wavelength pattern of radiation emitted by a source.
发射光谱是辐射源发出的辐射的波长模式。
Energy level is an allowed energy state in a model.
能级是模型中允许的能态。
An emitted photon corresponds to a transition to a lower energy level.
发射的光子对应向较低能级的跃迁。
Absorption requires a compatible energy difference.
吸收需要匹配的能量差。
Rutherford scattering supported a small dense nucleus, but that experiment alone did not establish the complete quantum model.
卢瑟福散射支持了小而致密的原子核，但仅凭该实验本身并未确立完整的量子模型。
In the lab: Read a labelled energy-level diagram before calculating.
在实验室中，计算前先阅读带标注的能级图。
Keep joules and electronvolts distinct and use the given constants.
将焦耳和电子伏特区分开，并使用给定的常数。
Compare attributed spectra at a common wavelength scale and avoid looking at unsafe light sources.
在共同的波长标度下比较所归属的光谱，并避免直视不安全的光源。
Known: energy levels are minus 6.0 and minus 2.0 electron volts.
已知：能级为负6.0和负2.0电子伏特。
A downward transition releases 4.0 electron volts, which equals 6.4 times 10 to the minus 19 joules, using 1 electron volt equals 1.6 times 10 to the minus 19 joules.
向下跃迁释放4.0电子伏特，即6.4乘以10的负19次方焦耳，其中1电子伏特等于1.6乘以10的负19次方焦耳。
With h equals 6.4 times 10 to the minus 34 joule seconds for this rounded exercise, frequency is 1.0 times 10 to the 15 hertz.
取h为6.4乘以10的负34次方焦耳秒进行本近似计算，频率为1.0乘以10的15次方赫兹。
Your turn: Levels are −7.0 and −2.0 eV.
轮到你了。 能级分别为负7.0和负2.0电子伏特。
Find the energy emitted in the downward transition.
求向下跃迁时释放的能量。 请暂停视频，自己算一算。
Pause the video and work it out. Check: the gap between the levels, minus 2.0 minus minus 7.0, is 5.0 electronvolts.
核对：两能级之差，−2.0 − (−7.0)，得 5.0 电子伏特。
Negative bound-state energies are relative to a chosen zero.
负的结合态能量是相对于所选零点而言的。
They do not mean that a negative photon energy is emitted.
这并不意味着会发射出负能量的光子。
A larger downward energy difference gives higher frequency.
向下的能级差越大，频率越高。
Not true: A negative bound-state energy means an emitted photon has negative energy.
不正确。 负的束缚态能量并不意味着发射的光子具有负能量。
Level energies are measured relative to a chosen zero (a free electron at rest), so bound states are negative.
能级能量是相对于选定的零点测量的，该零点是一个静止的自由电子，因此束缚态为负值。
The photon energy is the positive DIFFERENCE between levels - differences of negative numbers are positive.
光子能量是两个能级之间的正值差。 负数之差为正数。
Atoms have a small positive nucleus and electrons.
原子有一个小的带正电的原子核和电子。
An emitted photon corresponds to a transition to a lower energy level.
发射的光子对应于向较低能级的跃迁。
Watch for negative bound state energies are relative to a chosen zero; they do not mean a negative photon energy is emitted.
注意：负束缚态能量是相对于选定的零点而言的；这并不意味着会发射出负能量的光子。
Levels are minus 9.0 and minus 3.0 electronvolts.
能级分别为负9.0和负3.0电子伏特。
Find the photon energy of the downward transition.
求向下跃迁的光子能量。
A negative level does not make the photon energy negative.
负的能级不会使光子能量变为负值。
The result is a photon energy of 6 electronvolts.
结果是光子能量为6电子伏特。
An electron in a higher level drops to a lower one and emits a single photon.
处于较高能级的电子跃迁到较低能级，发射一个光子。
Its energy is exactly the difference between the two levels, so each transition gives one spectral line.
光子能量恰好等于两个能级之差，所以每次跃迁对应一条谱线。
The levels are minus 8, minus 3 and minus 1 electron volts.
三个能级分别为 −8、−3 和 −1 电子伏。
List all downward photon energies and identify the shortest wavelength.
列出所有向下跃迁的光子能量，并指出波长最短的一种。
State why these are discrete candidates.
说明为什么这些是分立的候选值。
The differences are 5, 7 and 2 electron volts; 7 electron volts has the shortest wavelength because E equals h c over lambda.
能量差分别为 5、7 和 2 电子伏；7 电子伏对应的波长最短，因为 E 等于 hc/λ。
Selection rules and populations may suppress a candidate transition.
选择定则和能级布居可能使某个候选跃迁不出现。
Pause and check yourself. Explain each point on screen in your own words, then change one number in the last example and say which assumption must still hold.
暂停一下，自我检查：用自己的话解释屏幕上的每个要点，再改动最后一个例子中的一个数字，说明哪个假设仍然必须成立。