Question #169200

 Three spin less non-interacting distinguishable particles, with respective masses  π‘š1, π‘š2, and π‘š3 in the ratio π‘š1: π‘š2: π‘š3 = 1: 2: 3, are subject to a common infinite  square well potential of width L in one spatial dimension. Determine the energies and the corresponding wave functions in the lowest two energy states of the system.


1
Expert's answer
2021-03-08T08:26:26-0500

Answer

common infinite square well potential of width L in one spatial dimension

Energy is given

En=n2Ο€2ℏ22mL2E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}

Wavefunction

ψ=2Lsin(nΟ€xL)\psi=\sqrt{\frac{2}{L}}sin(\frac{n\pi x}{L})

According to given data total energy of system is given

For ground state n=1

E1=n2Ο€2ℏ22m1L2+n2Ο€2ℏ22m2L2+n2Ο€2ℏ22m3L2E_1=\frac{n^2\pi^2\hbar^2}{2m_1L^2}+\frac{n^2\pi^2\hbar^2}{2m_2L^2}+\frac{n^2\pi^2\hbar^2}{2m_3L^2}

E1=E_1= 7Ο€2ℏ28mL2\frac{7\pi^2\hbar^2}{8mL^2}


First excited state n=2

E2=n2Ο€2ℏ22m1L2+n2Ο€2ℏ22m2L2+n2Ο€2ℏ22m3L2E_2=\frac{n^2\pi^2\hbar^2}{2m_1L^2}+\frac{n^2\pi^2\hbar^2}{2m_2L^2}+\frac{n^2\pi^2\hbar^2}{2m_3L^2}

=E1=7Ο€2ℏ22mL2E_1=\frac{7\pi^2\hbar^2}{2mL^2}

Now wavefunction

ψ1=2Lsin(Ο€xL)\psi_1=\sqrt{\frac{2}{L}}sin(\frac{\pi x}{L})

And

ψ2=2Lsin(2Ο€xL)\psi_2=\sqrt{\frac{2}{L}}sin(\frac{2\pi x}{L})


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