Explain the nature of escape peaks in gamma spectroscopy with a HP Ge detector. How does their efficiency depend on the geometry of the detector?
Answer:
In Gamma Spectrum with HP Ge detector, there are two kinds of escape peaks:
1. Single Escape Peak
2. Double Escape Peak
So, the Single Escape Peak appears due to the escape of one of the 511KeV gamma ray produced in the pair production and it appears at (E-511) KeV as shown in figure below in red mark.

The double Escape Peak appears due to escape of both of the 511KeV gamma ray produced in the pair production and it appears at (E-511) KeV as shown in figure below in red mark:

The peak shape is Gaussian in nature, which when differentiated two times i.e. when its second derivative is taken, gives a central large negative peak flanked by two smaller positive peaks. Whereas a smooth background gives, a constant first derivative and hence a zero second derivative and a Compton edge gives only one positive peak instead of two peaks as in a Gaussian peak. This behavior of the second derivative is considered as the signature of the peak and it is located at the center of the negative lobe. If there are two or more close by peaks than the behavior will be still different from a single peak.
Part B:
It depends by a factor of 4 pi.


Explain the nature of escape peaks in gamma spectroscopy with a HP Ge detector. How does...
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113 pts) consider the molecules shown below. Briefly explain how you could use IR spectroscopy to distinguish between each pair of compounds. Give approximate frequencies for any diagnostic peaks you would expect (or not expect) to observe. Focus on the key signal(s) that would allow you to tell the two apart.
For
IR spectroscopy assigning peaks, could you please explain to me how
do we tell which structure belong to each regions of peaks.
or C -H 000 3800 30 3400 3200 3000 2800 2000 2400 2200 C-H stretch of ep Alkene 2 9 10 5 26 2.7 28 293 2000 4000 3800 3600 3400 3200 3000 2800 2600 2400 2200 Alkene 3 0r 26 27 28 29 3 NICOLET sx 4000 soo 3500 300 3200 3000 2to0 2000 N 0...
From the infrared spectroscopy and C13 NMR
explain how this proves its malonic acid.
the image with 2 peaks is C13 NMR.
170.938 40.513 200 150 F1: 15.092 EX: CZ.DE SWT: 5000 PW:9.7. OE 1: 152LO NA: 12 PTSI ISI 10/23/19 04:26) PD: 3.7 se !%TIL !!! 65 4000 1500 3500 2500 3000 2500 2000 1000 450 cm-1
Explain (in your own chemically accurate words) why and how you can use IR spectroscopy to measure bonding parameters for a polar, diatomic molecule. You will need to address why the ro-vibrational spectrum is in the IR region of the electromagnetic spectrum (this may include a discussion of vibrational and rotational motion and the selection rules associated with them) the origin of the P, Q and R branch (including a figure to indicate the origin of each set of peaks...
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