求解🙏

The photoelectric effect can be used to measure the value of Planck's constant. NOTE: The following data are fictional, and will give a result that is quite different from the real value of Planck's constant. Be sure that you do not use the real value of Planck's constant in any calculations here. 1 S -1 Suppose that a photoelectric effect experiment was carried out using light with v = 7.56 × 10¹4 s-¹, and ejected electrons were detected with a kinetic energy of 2.41 × 10-¹1 J. The experiment was then repeated using light with = 9.81 × 10¹4 s and the same metal target, and electrons were ejected with a kinetic energy of 9.09 × 10-¹1 J. Use these data to find a value for Planck's constant (in J • s). Hint. It may help to start by thinking about how you would calculate the metal's binding energy if you already knew Planck's constant. Submit Answer S Try Another Version 2 item attempts remaining

1.27要怎麼解是代n=ni 平方/NA?

1.27 For a silicon crystal doped with boron, what must NÅ be if at T = 300 K the electron concentration drops below the intrinsic level by a factor of 10%? Ans. N₁ = 1.5 × 10¹/cm³ 1.9 Current Flow in Semica . 10 58 af bill yd basilma

請問第39（1）怎麼計算⋯？🙏😭

TOOTH tools to Across a thin cell membrane, the solution pH changes from 7.2 to 5.2, Calculate the electric potential difference across the membrane. Ans.0.59210 Juoi blogs (2) Which side (7.2 side or 5.2 side) has a higher potential? ole (1)0.11832V; obri. (2) pH = 5.2 VI VARI 12 Lynsm wol/25 (103台大A) osups mis 5.2

想請問這個表格上寫的 "fold reversal "是什麼意思 要怎麼翻譯比較好，謝謝（有上網查過了，但不太了解）

Pro DC 箱(W) 說明(H) nib).pdf x ↑ X gs 638 (3/14) 125% ... Table 1. Effect of lazertinib on reversing the MDR mediated by ABCB1 or ABCG2 in drug selected resistant cancer cells IC 50 ± SD (µM) (fold reversal) Compounds KB (parental) KBv200 (ABCB1) Vincristine 0.0059 +0.0323 (1.00) 0.2263 ± 0.0323 (1.00) +0.0625 µM Lazertinib 0.0053 ± 0.0003 (1.12) 0.0041 ± 0.0002 (1.43) 0.0782 +0.0182** (2.89) 0.0195 ±0.0105** +0.125 µM Lazertinib (11.60) 0.0090 0.0011** (25.21) +0.25 µM Lazertinib 0.0050 +0.0007 (1.54) 0.0042 ± 0.0002 (1,47) 0.0100 ± 0.0038** (22.60) +10 µM Verapamil Doxorubicin 0.0218 ± 0.0038 (1.00) 1.2091±0.4592 (1.00) +0.0625 µM Lazertinib 0.0216 ± 0.0093 (1.01) 0.4347±0.0807** (2.76) +0.125 uM Lazertinib SD alo and *p exp tor res an apy ure wh ing E Confe 100% M C 從表1年 對受質車 ABCG2 然而·L= 劑的IC5C 藥物敏感

大學普化 方智化學第六題

ow Na k 7 CI o se Br Chapter 2 化學鍵結 AH -C=CH 升二技插大 3. Brigma The bond between the carbon atoms in acetylene (HCCH) consists of ) (A)6 pi electrons (B)4 pi and 2 sigma electrons (C)6 sigma electrons (D)2 pi and 4 sigma electrons (E)3 pi and 3 sigma B) B 7.(BCDE) electrons. 【84 成大化學) 14.(B) 20.(C) 4. E ) (A)NO; [83 中興A] Which of the following has a triple bond? (B)O2 (C)C12 (D)CO; (E)CN-. Which one of the following element pairs has the lowest electronegativity (B) 27.(D) 34. (C) C) 41.(CD) 1 G difference and thereby is the least polan? Få 10 323.49 B) (84 5% *ILI) -) 48. (A) 6. NET 2 > 2 D) 58. (D) 62. (B) 69. NO+ D Obind (219) O 发 (A)SeF (B)Seci (QKBr (D/NaCl (E)KAT. 11 The compound Sio does not exist as a discrete molecule while CO2 does. This can be explained because ly 2 ep 3 15 aly zp (A)the Si–O bond is unstable (B)The Lewis structure of SiO has an even number of electrons (C)The SiO2 is a solid while CO, is a gas (D)the 3p orbital of the Si has little overlap with the 2p of the 0 (E)none of the these. V MOJT 515 1 - 4 40 (成大】 Yi (4 0 atom) (86 58*A] 7. In order to explain the valence observed for many of the elements 0 -Si-O in terms of electron configurations, it is often useful to look at electronis staies slightly different from the ground state in comparing 0-41-0 the ground state of boron with the state more useful to describe 0 bonding, the number of unpaired electrons goes from (A)1 to 2 (B)1 to 3 (C)3 to 4 (D)3 to 5 (E)none of the above. (集 (80 清大B) Which of the following compounds should not exist? ( (, 【85 成大A】 alf 566-32 Cle 32-8-24 2-89 cild 14 1 he atoms 83 中興A) mm bove. B 29 p. 1 1 be a ) \CHE > ? (A)Na,P (B)(NH4),PO, CIPO. DPH xH)POCI,, Maspal sput 中興A] Pa zorgt O 15412 = 1) O-P- 5 5+12= 1) 10-4 - 13 13-12=1 -- cl-P-0

求詳解謝謝各位🙏

28. Which one of the following will have the greatest number of unpaired electrons ? A. B. C. [Mn(H2O).]2+ (weak field) [Fe(en)3]3+ (strong field) [Ru(NH3).]?* (strong field) (Cr(CI).]*- (weak field) [Fe(en)3]** (strong field) D. E.

有人知道我是哪裡算錯嗎(ಥ_ಥ)

= 78, When the electron in a hydrogen atom falls from the n= 3 excited energy level to the ground state energy level, a photon with wavelength 2 is emitted. An electron having x this same wavelength would have a velocity of A) 7.10 x 10 m/s. D) B) 2.93 x 10 m/s. E) C) 2.93 x 103 m/s. Ans: A Category: Medium Section: 7.4 7.10 m/s. 3.00 x 108 m/s.

請問第68題要怎麼看

Ans: B Section: 7.9 75. Category: Medium E 66. A ground-state chromium atom has how many unpaired electrons? (58) 24 A) 1 B) 2 C) 4 D) 5 E) 6 IS2537355P 3843 Category: Medium Section: 7.9 Ans: E 16. atom? 67. Which of the following is the electron configuration of an excited state of an oxygen C A) 1s22s22p4 B) 1s 2s²2p" C) 1822s22p%35' D) 1s22s22p E) 1s22s22p3 Category: Medium Section: 7.9 Ans: C 68. Which of the following is the electron configuration of an excited state of an iron atom? A) [Ar]45²3d? B) [Ar]4s 3d C) [Ar]4s3d8 D) [Ar]4s'3d? E) [Ar]4s 3d Ans: D Category: Medium Section: 7.9 7 CM 1453

為什麼不是E

Q搜尋 07.pdf Chapter 7: Atomic Structure and Periodicity 1. When ignited, a uranium compound burns with a green flame. The wavelength of the light given off by this flame is greater than that of A) red light B) infrared light C) radio waves D) ultraviolet light E) gamma rays ANS: D PTS: 1 DIF: Easy REF: 7.1 KEY: Chemistry | general chemistry | atomic theory light | electromagnetic radiation MSC: Conceptual 2. Which form of electromagnetic radiation has the longest wavelengths? A) gamma rays

求解答案是否正確

Multiple-choice questions: (70 points) write the letter of statement that is not true: 1. A.B.U The Drude Model: (a) It is a classical approach using concepts from the kinetic theory of gases. (b) A relaxation-time approximation is used where the probability of collision is l/t. (c) No forces act on the electrons in between collisions. (d) It predicts accurately the electronic heat capacity of metals. 2. The Sommerfeld Model: (a) The crystal potential is replaced by an average Coulomb potential. (b) The separation of states is so close that a continuum model can be applied. (c) The electrons are highly localized due to a strong crystal potential. (Q) The energy of the highest occupied state at 0 K is the Fermi energy. 3. Density of states in k-space, N(k): (a) N(k) is determined by the boundary value conditions. (b) For lattices in any dimensions, N(k) is constant in k-space. (C) N(k) depends on the direction and magnitude of k. (d) It is directly proportional to the volume of the crystal. E影響, c 4. The Fermi surface of a metal: (a) is responsible for many of the transport properties of the metal, (b) depends on the nature of the chemical bonds, (c) is always a sphere, (d) can involve mixing of electron orbitals. ✓ 5. The conductivity of a material () increases with the mobility of its carriers, (b) increases with the density of carriers, (C) it is inversely proportional to the energy gap. cogu changes with temperature. 6. BID Plasmons: (a) They are an important characteristic of the Sommerfeld electron gas model. (b) Plasmons occur spontaneously in metals. (c) They are related to charge density oscillations. (d) The Plasmon energy is typically of the order of 10 eV. The Fermi energy Ef in intrinsic semiconductors 8) is usually close to y E, at OK (b) it ACiD. can increase or decrease with temperature depending on the ratio of hole and electron effective masses, (c) it represents the point at which the probability of occupation is 12, even though there may not been any states with that energy, (d) its value relative to the electronic bands must remain unaffected in the presence of defects or interfaces.