A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en

Question

2 months ago

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ergy of 6.0 eV. What is the energy of this particle in the ground state?

1 answer:

kicyunya [3.2K] · Answer 1 · 2026-03-25T12:36:38+03:00

Answer:

The particle's energy in its ground state is E₁=1.5 eV.

Explanation:

For a particle with mass m in the nth energy level of an infinite square well potential of width L , the energy $E_{n}$ is given by:

$E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}$

In the ground state (n=1) and in the first excited state (n=2), where the energy is noted as E₂= 6.0 eV. Substituting into the above equation yields:

$E_{1}=\frac{h^{2}}{8mL^{2}}$

$E_{2}=\frac{h^{2}}{2mL^{2}}$

Thus, we can express the ground state's energy as:

$E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})$

$E_{1}=\frac{1}{4} E_{2}$

$E_{1}=\frac{1}{4} ( 6.0\ eV)$

Ultimately

$E_{1}=1.5\ eV$