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CdSe-based Semiconductor Nanocrystal: Synthesis, Characterization, and Applications

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𝐸 = 𝐸𝑔+ 2

8𝑅2 ( 1

𝑚𝑒+ 1

𝑚) −1.8𝑒2

4𝜋𝜀𝑅

𝐸𝑔

ℎ 𝑚𝑒 𝑚

𝑅 𝜀

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χ

I(t) = ∫ ρ(Γ) · e0 −iΓtdt ρ Γ 𝜌(Γ) = 1

𝜎Γ√2𝜋𝑒

1 2(lnΓ−𝜇

𝜎 )2

Γ σ

Γ Γ

τ = 1 𝛤⁄ 𝑚𝑓

Γ Γ

τ

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τ

Γ τ Γ

Γ

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𝐷(𝑛𝑚) = 59.60816 − 0.54736𝜆 + 1.8873𝑥10−3𝜆2− 2.85743𝑥10−6𝜆3+ 1.62974𝑥10−9𝜆4 = 2.5 𝑛𝑚

𝜆

𝛥𝐸1𝑆,𝐹𝑊𝐻𝑀 (𝐸1𝑆

λ

(173)

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𝐸1𝑆 (𝑒𝑉) = ℎ (eV s−1) 𝑥 𝑐(𝑚s−1)

𝜆 (𝑚) = 2.4486 𝑒𝑉(AI. 2)

ℎ 𝑐 𝜆

𝜀1𝑆 𝛥𝐸1𝑆,𝐹𝑊𝐻𝑀

𝐸1𝑆 𝑙

[𝐶𝑑𝑆𝑒] = 𝐴𝑏𝑠

𝑙(𝑐𝑚)𝜀1𝑆(𝑀−1cm−1)𝑥𝛥𝐸1𝑆,𝐹𝑊𝐻𝑀

0.06 = 2.4 𝑥10−6𝑀(AI. 3)

𝑃𝑀 (𝑑𝑜𝑡)

𝑃𝑀 (𝑑𝑜𝑡) ∶ 𝑑(𝑛𝑚)

𝑙𝑎𝑡𝑡𝑖𝑐𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡(𝐶𝑑𝑆𝑒) = 2.5

0.35= 7.1428 (AI. 4)

𝐶𝑑𝑆𝑒 𝑢𝑛𝑖𝑡𝑠 − 𝑛𝑢𝑚𝑏𝑒𝑟 𝑝𝑒𝑟 𝑑𝑜𝑡: 4

3𝑥𝜋𝑥(𝑃𝑀(𝑑𝑜𝑡)

2 ) 3 = 190.71(AI. 5)

𝑃𝑀

𝐷𝑜𝑡 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠: 𝐶𝑑𝑆𝑒 𝑢𝑛𝑖𝑡𝑠 𝑥 𝑃𝑀(𝐶𝑑𝑆𝑒) = 190.71 𝑥 191.37

= 36496.1727(AI. 6)

(174)

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𝑛 𝑑𝑜𝑡𝑠 ∶ [𝐶𝑑𝑆𝑒] 𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 = 2.4 𝑥10−6𝑥 2 𝑥10−3

= 4.8 nmol(AI. 7)

𝑇𝑜𝑡𝑎𝑙 𝐶𝑑𝑆𝑒 𝑚𝑎𝑠𝑠: 𝑛 𝑑𝑜𝑡𝑠 𝑥 𝐷𝑜𝑡 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠

= 4.8𝑥10−9𝑥 36496.1727 = 1.7518𝑥10−4𝑔(AI. 8) 𝑁𝐴

𝐶𝑑𝑆𝑒 𝑚𝑎𝑠𝑠 𝑝𝑒𝑟 𝑑𝑜𝑡: 𝐷𝑜𝑡 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠

𝑁𝐴 = 36496.1727

6.023𝑥1023

= 6.05946𝑥10−20𝑔(AI. 9)

𝐷𝑜𝑡𝑠 𝑛𝑢𝑚𝑏𝑒𝑟: 𝑇𝑜𝑡𝑎𝑙 𝐶𝑑𝑆𝑒 𝑚𝑎𝑠𝑠

𝐶𝑑𝑆𝑒 𝑚𝑎𝑠𝑠 𝑝𝑒𝑟 𝑑𝑜𝑡 𝐷𝑜𝑡 = 1.7518𝑥10−4 6.05946𝑥10−20

= 2.8910𝑥1015𝑁𝐶𝑠(AI. 10)

𝑉𝑖

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑁𝐶𝑠 𝑣𝑜𝑙𝑢𝑚𝑒 (𝑉𝑖): 4

3 𝑥 𝜋 𝑥 𝑟𝑖3 = 8.177 𝑛𝑚3(AI. 11) 𝑑𝑓

𝑑𝑖 𝑎

𝑑𝑓: 𝑑𝑖 + 𝑎(𝑍𝑛𝑆) + 𝑎(𝑍𝑛𝑆) = 2.5 + 0.541 + 0.541

= 3.582 𝑛𝑚(AI. 12)

(𝑉𝑓) 𝐹𝑖𝑛𝑎𝑙 𝑁𝐶 𝑣𝑜𝑙𝑢𝑚𝑒𝑛 (𝑉𝑓): 4

3 𝑥 𝜋 𝑥 𝑟𝑓3 = 24.052 𝑛𝑚3(AI. 13)

(175)

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𝑉(𝑍𝑛𝑆) = 𝑉𝑓 – 𝑉𝑖 = 15.8752 𝑛𝑚3(AI. 14)

𝑍𝑛𝑆 𝑐𝑒𝑙𝑙 𝑛𝑢𝑚𝑏𝑒𝑟: 𝑉(𝑍𝑛𝑆)

𝑉(𝑍𝑛𝑆𝑢𝑛𝑖𝑡𝑐𝑒𝑙𝑙) = 15.8752 0.1583

= 100.28 𝑐𝑒𝑙𝑙𝑠(AI. 15)

𝑚(𝑍𝑛𝑆 )𝑝𝑒𝑟 𝑑𝑜𝑡: 𝑚(𝑍𝑛𝑆) 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 𝑥 𝑐𝑒𝑙𝑙𝑠 𝑛𝑢𝑚𝑏𝑒𝑟

= 6.32𝑥10−22𝑥 100.28

= 6.338𝑥10−20𝑔 𝑜𝑓 𝑍𝑛𝑆(AI. 16)

𝑇𝑜𝑡𝑎𝑙 𝑍𝑛𝑆 𝑚𝑎𝑠𝑠 𝑛𝑒𝑐𝑒𝑠𝑠𝑎𝑟𝑦: 𝑚(𝑍𝑛𝑆)𝑝𝑒𝑟 𝑑𝑜𝑡 𝑥 𝑑𝑜𝑡 𝑛𝑢𝑚𝑏𝑒𝑟

= 6.338𝑥10−20𝑥 2.8910𝑥1015

= 1.832𝑥10−4𝑔(AI. 17)

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Φ𝐹 = 𝑁𝑜. 𝑜𝑓 𝑒𝑚𝑚𝑖𝑡𝑒𝑑 𝑝ℎ𝑜𝑡𝑜𝑛𝑠

𝑁𝑜. 𝑜𝑓 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 𝑝ℎ𝑜𝑡𝑜𝑛𝑠

⁄ (AII. 1)

Φ𝐹(𝑋) = (𝐴𝑆 𝐴𝑋

⁄ (𝐹𝑋 𝐹𝑆

⁄ (𝑛𝑋 𝑛𝑆

⁄ )2 𝛷𝐹(𝑆)(AII. 2) Φ𝐹(𝑋)

(182)

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φ

(183)

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𝜑 =𝐸𝑐 − 𝐸𝑎

𝐿𝑎− 𝐿𝑐 (AII. 3)

(184)

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λ

χ

I(t) = ∫ ρ(Γ) · e0 −iΓtdt (AII. 4) ρ Γ

𝜌(Γ) = 1

𝜎Γ√2𝜋𝑒

1 2(lnΓ−𝜇

𝜎 ) 2

(AII. 5)

Γ σ

Γ Γ

𝛤𝑚𝑓 = 𝑒𝜇−𝜎2(AII. 6) 𝛥𝛤 = = 2 · 𝛤 · 𝑠𝑖𝑛ℎ(𝜎)(AII. 7)

τ = 1 𝛤⁄ 𝑚𝑓(AII. 8)

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Γ Γ

λ

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α

 

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ν

ν

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δ

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α

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ρ ρ

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(𝑉𝑜𝑙)2 × (𝜌 − 𝜌𝑠𝑜𝑙𝑣𝑒𝑛𝑡)2 × 𝑁 = 𝐹(AII. 9)

ρ ρ

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