The band gap in the Aluminium antimonide solid state laser for wavelength of 730 nm has to be calculated. Concept introduction: According to Band theory of solids, the energy levels of a substance are imagined as “bands”. There are two types of bands – valence band and conduction band. Low lying bands are valence band and conduction band where the conduction takes place, lies above the valence band. The energy gap between the valence band and conduction band is called “band gap”. The energy gap can be represented by Planck’s equation, E = hν where ν = c λ E = energy h = Planck's constant ν = frequency c = velocity of light λ = wavelength
The band gap in the Aluminium antimonide solid state laser for wavelength of 730 nm has to be calculated. Concept introduction: According to Band theory of solids, the energy levels of a substance are imagined as “bands”. There are two types of bands – valence band and conduction band. Low lying bands are valence band and conduction band where the conduction takes place, lies above the valence band. The energy gap between the valence band and conduction band is called “band gap”. The energy gap can be represented by Planck’s equation, E = hν where ν = c λ E = energy h = Planck's constant ν = frequency c = velocity of light λ = wavelength
Solution Summary: The author explains that the band gap in the Aluminium antimonide solid state laser for wavelength of 730 nm has to be calculated.
The band gap in the Aluminium antimonide solid state laser for wavelength of 730 nm has to be calculated.
Concept introduction:
According to Band theory of solids, the energy levels of a substance are imagined as “bands”. There are two types of bands – valence band and conduction band. Low lying bands are valence band and conduction band where the conduction takes place, lies above the valence band. The energy gap between the valence band and conduction band is called “band gap”. The energy gap can be represented by Planck’s equation,
Arrange the solutions in order of increasing acidity. (Note that K (HF) = 6.8 x 10 and K (NH3) = 1.8 × 10-5)
Rank solutions from least acidity to greatest acidity. To rank items as equivalent, overlap them.
▸ View Available Hint(s)
Least acidity
NH&F NaBr NaOH
NH,Br NaCIO
Reset
Greatest acidity
1. Consider the following molecular-level diagrams of a titration.
O-HA molecule
-Aion
°°
о
°
(a)
о
(b)
(c)
(d)
a. Which diagram best illustrates the microscopic representation for the
EQUIVALENCE POINT in a titration of a weak acid (HA) with sodium.
hydroxide?
(e)
Answers to the remaining 6 questions will be hand-drawn on paper and submitted as a single
file upload below:
Review of this week's reaction:
H₂NCN (cyanamide) + CH3NHCH2COOH (sarcosine) + NaCl, NH4OH, H₂O --->
H₂NC(=NH)N(CH3)CH2COOH (creatine)
Q7. Draw by hand the reaction of creatine synthesis listed above using line structures without showing
the Cs and some of the Hs, but include the lone pairs of electrons wherever they apply. (4 pts)
Q8. Considering the Zwitterion form of an amino acid, draw the Zwitterion form of Creatine. (2 pts)
Q9. Explain with drawing why the C-N bond shown in creatine structure below can or cannot rotate. (3
pts)
NH2(C=NH)-N(CH)CH2COOH
This bond
Q10. Draw two tautomers of creatine using line structures. (Note: this question is valid because problem
Q9 is valid). (4 pts)
Q11. Mechanism. After seeing and understanding the mechanism of creatine synthesis, students should
be ready to understand the first half of one of the Grignard reactions presented in a past…