Question

a particle is moving in a three-dimensional potential V(x,y,z) = (1/2)mw²(2x²+y²+4z²). If the mass of the particle is m, then what is the energy of the particle in the lowest state?​

Answer 1

Answer:

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Explanation:

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Answer 2

Step-by-step Explanation:

Given: Three-dimensional potential $V(x,y,z) = \Big{\frac{1}{2}} \ m \omega^{2} (2x^{2} +y^{2} +4z^{2} )$

To Find: The energy of the particle in the lowest state

Solution:

• Finding energy of the particle in the lowest state

We have a three-dimensional potential $V(x,y,z) = \Big{\frac{1}{2}} \ m \omega^{2} (2x^{2} +y^{2} +4z^{2} )$

such that we can write,

$\Rightarrow V(x,y,z) = \Big{\frac{1}{2}} \ m (2\omega^{2}) x^{2} + \Big{\frac{1}{2}} \ m (\omega^{2})y^{2} + \Big{\frac{1}{2}} \ m(4\omega^{2})z^{2}$

Replacing the term $2\omega^{2} \rightarrow \omega^{2}_{x}$, $\omega^{2} \rightarrow \omega^{2}_{y}$, and $4\omega^{2} \rightarrow \omega^{2}_{z}$ such that

$\Rightarrow V(x,y,z) = \Big{\frac{1}{2}} \ m x^{2} \omega^{2}_{x} + \Big{\frac{1}{2}} \ m y^{2} \omega^{2}_{y} + \Big{\frac{1}{2}} \ mz^{2} \omega^{2}_{x}$

The above expression is the potential of a three-dimension harmonic oscillator. Therefore, the Eigen energy value for the harmonic oscillator is

$E_{n_{x}n_{y}n_{z}} = ({n_{x}} + \frac{1}{2} ) \hbar \omega_{x} + ({n_{y}} + \frac{1}{2} ) \hbar \omega_{y} + ({n_{z}} + \frac{1}{2} ) \hbar \omega_{z}$

at the lowest energy state, $n_{x} = n_{y} = n_{z} = 1$. Thus, we can write;

$E_{n_{x}n_{y}n_{z}} = (1 + \frac{1}{2} ) \hbar (\sqrt{2} \omega) + (1 + \frac{1}{2} ) \hbar (\omega) + (1 + \frac{1}{2} ) \hbar (2\omega)$

$\Rightarrow E_{1,1,1} = \frac{3}{2} \hbar \omega (\sqrt{2} + 1 +2 )$

$\Rightarrow E_{1,1,1} = \frac{3}{2}(\sqrt{2} + 3 ) \hbar \omega$

Hence, the energy of the particle in the lowest state is $\frac{3}{2}(\sqrt{2} + 3 ) \hbar \omega$