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jarptica [38.1K]
3 years ago
13

I'm in  my lab, writing an essay about the properties of Bose Einstein Condensate. I forgot the formula of producing the condens

ate. Can someone help me?
Physics
1 answer:
Artyom0805 [142]3 years ago
7 0
This is a big looking question for school and for 5 points.It seems to relate to superfluidity in helium III and helium IV and something called the "lambda" point. Though I can't do it justice yet and here, it looks as though i can find tis in "Heat and Thermodynamics", Mark W Zemansky ...If rho is the density of helium II, rhon the density of the normal part, and rhoz the density of the superfluid part rho=rhon+rhoz.At the lambda point, all the atoms are normal and rhon/rho = 1, whereas at absolute zero all the atoms are superfluid and rhon/rho=0.Best I can do with the info and point available ..ps, I think that both Bose and Einstein were nobel prize winners, and the word "Boson" is in honour of the, I believe, Indian physicist Bose. It is a very interesting question, and I can in no way do it justice here.
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3 years ago
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Explanation:

Given that,

(a) Speed, v=6.66\times 10^6\ m/s

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The wavelength of the electron is given by :

\lambda_e=\dfrac{h}{m_ev}

\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}

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\lambda_p=\dfrac{h}{m_p v}

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\lambda_p=5.96\times 10^{-14}\ m

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The relation between the kinetic energy and the wavelength is given by :

\lambda_e=\dfrac{h}{\sqrt{2m_eK}}

\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}

\lambda_e=1.18\times 10^{-11}\ m

\lambda_p=\dfrac{h}{\sqrt{2m_pK}}

\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}

\lambda_p=2.77\times 10^{-13}\ m

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Answer:

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