I am looking forward to depositing Cu and Ni thin films on Si/SiO2 substrates. Is it possible to deposit Cu and Ni thin films by using the thermal evaporation technique. ?
Copper without problems. Nickel must be tasted. Usually we deposit nickel films with an electron beam method. But it is possible that the tungsten vaporizer will be able to precipitate by the thermal method. If you need help, please contact. We will be glad
You can do this if the thermal evaporation boat is made of tungsten or molybdenum to evaporate both of the aforementioned elements so that the distance between the sample boat and the holder of the substrates to be vaporized on is 17 cm. It is the best distance for thermal evaporation.
cupper is very easy to deposit as thin film by thermal evaporation employing molybdenum boat ,but Ni is also possible under certain conditions
thank you so much for information and instant responses Prof. Sergei Petrushenko, Prof. Ammar Qasem's, Prof. Nadir Fadhil Habubi and Dr. Kashinath Lellala
yes.that is possible.but that is very difficult and this work needs to high attention.
It is possible to use thermal evaporation technique to deposit both Cu and Ni on your substrate.
Yes. The first you can Nickel metal with high purity deposited on silicon substrate by thermal evaporation method and then high purity copper metal is deposited on top of the Ni layer. The thickness of the nickel layer is about 30-45 nm and the thickness of Cu on Ni is about 25-35 nm.
Thank you Dr. Noori for your advice and suggestion to deposit metal films by thermal evaporation. If you have any literature on the deposition of Cu/Ni thin film by thermal evaporation, can you please share it ?
We evaporate nickel all the time using thermal evaporation. We use WBAO-2 alumina-coated tungsten baskets from R. D. Mathis (also available from other suppliers) at around 35 A and 7 V. The deposition rate is in the range of 4 Å/s at a distance of 8 cm.
Theoretically:
From XRD patterns as follows:
Variations in the height of the specimen in its holder, whether increasing or decreasing, cause XRD peaks to change from their original location.
Important and useful pathways are connected to the XRD Curve:
Scherrer’s equation:
Particle Size, D = (0.9 x λ)/ (d cosθ)
λ = 1.54060 Å (in the case of CuKa1) so, 0.9 x λ = 1.38654
Θ = 2θ/2 (in the example = 20/2)
d = the full width at half maximum intensity of the peak (in Rad) – you can calculate it using Origin software.
To convert from angle to rad
Rad = (22 x angle) / (7 x 180) = angle x 0.01746
Example: if d = 0.5 angle (θ)
= (22 x 0.5)/ (7x 180) = 0.00873 rad
Then
Delta=1/D^2
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