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dusya [7]
3 years ago
7

Bosons & Fermions A new type of quantum object has been discovered. When a large collection of these are cooled to almost ab

solute zero, they are observed to all condense into the same state - the ground state. Select all that are True.
Is this object a Fermion obeying Fermi-Dirac statistics?
Is this object a Boson obeying Bose-Einstein statistics?
Could this object have intrinsic spin 2 hbar?
Would the Schrodinger equation for two of these particles be symmetric with exchange of the two particles?
Could this object have intrinsic spin s = 3/2 hbar?
Would the Schrodinger equation for two of these particles be anti-symmetric with exchange of the two particles?
Physics
1 answer:
Lena [83]3 years ago
5 0

Answer:

Only 2,3,4 are true

Explanation:

Bosons Particles are particles that condense to the same state. Bosons particle have integral spin like 0 , $\hbar$, 2$\hbar, 3$\hbar$, etc. Bosons particles always have asymmetric wave function and there is exchange of particles.

1) It does not obey Fermi_ Dirac statistics

2) It obeys Bose-Einstein statistics

3) The object can have intrinsic spin 2$\hbar

4) Yes the Bosons particle is always symmetric with exchange of particles

5) No Bosons particle are symmetric and not asymmetric

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What are some actions that can be taken to remedy air pollution? Select all that apply. Run your air conditioner at a higher set
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4 0
3 years ago
An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place
anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

mu=(M+m)v

v=\frac{m}{M+m}\times u

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

\frac{(M+m)v^2}{2}=(M+m)gh

here h=L(1-\cos \theta ) from diagram

therefore

v=\sqrt{2gL(1-\cos \theta )}

initial velocity in terms of v

u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}

For first case \theta =6.8^{\circ}

u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}

for second case \theta =11.4^{\circ}

u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}

Therefore \frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}

\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}

\frac{u_1}{u_2}=1.084

i.e.\frac{v_1}{v_2}=1.084

4 0
3 years ago
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