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Chapter 6: Q68CP (page 1405)

Each of 2Nelectrons (mass m) is free to move along the x-axis. The potential-energy function for each electron is,

U(x)=12k'x2

where k_ is a positive constant. The electric and magnetic interactions between electrons can be ignored. Use the exclusion principle to show that the minimum energy of the system of 2Nelectrons is role="math" localid="1664001166050" hN2k'/m. (Hint:See Section 40.5 and the hint given in Problem 41.47.)

Short Answer

Expert verified

It is proved that theminimum energy of the system of 2Nelectrons is hN2k'/m

Step by step solution

01

(a) Identification of the concept. 

For a simple harmonic oscillator, the potential energy is,

Ux=12k'x2

Also, energy of each state is,

E=n+12hω

As per the Pauli’s Exclusion Principle, there can be two electrons in a state denoted by n as ms = +1/2 and -1/2. So, the quantum states are defined by the pair (n, ms).

02

(b) Determination of the minimum energy of the system of 2N electrons.

The minimum energy is calculated as,

2∑n=0N-1En=2∑n=0N-1n+12hω=2hω∑n=0N-1n+∑n=0N-112=2hωN-1N2+N2=hωN2-N+N=hωN2=hN2k'm

Minimum energy is hN2k'm.

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