Physics, 07.11.2019 20:31 sugeilochoa

An electron is trapped in an infinite square-well potential of width 0.4 nm. if the electron is initially in the n = 4 state, what are the various photon energies that can be emitted as the electron jumps to the ground state? (list in descending order of energy. enter 0 in any remaining unused boxes.)answers are, but cannot solve to ithighest: 35.3 ev28.2 ev18.8 ev16.5 ev11.8 evlowest: 7.06 ev

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Answer from: bertrandcherlan

The energies in descending order 35.3 eV, 28.2 eV, 18.8 eV, 16.5 eV, 11.8 eV, 7.05 eV.

Explanation:

Given that,

Width = 0.4 nm

Number of state n= 4

We need to calculate the energy in first state

When n = 1,

Using formula of energy

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

Put the value into the formula

$E_{1}=\dfrac{(6.63\times10^{-34})^2}{8\times9.1\times10^{-31}\times(0.4\times10^{-9})^2}$

$E_{1}=3.77\times10^{-19}\ J$

$E_{1}=\dfrac{3.77\times10^{-19}}{1.6\times10^{-19}}$

$E_{1}=2.35\ eV$

Energy in nth level is

$E_{n}=n^2\times E_{1}$

Photon energy in different states is

$E_{p}=E_{n}-E_{n-1}$

Put the value into the formula

$E_{4-3}=(4^2-3^2)\times2.35=16.5\ eV$

$E_{4-2}=(4^2-2^2)\times2.35=28.2\ eV$

$E_{4-1}=(4^2-1^2)\times2.35=35.3\ eV$

$E_{3-2}=(3^2-2^2)\times2.35=11.8\ eV$

$E_{3-1}=(3^2-1^2)\times2.35=18.8\ eV$

$E_{2-1}=(2^2-1^2)\times2.35=7.05\ eV$

Hence, The energies in descending order 35.3 eV, 28.2 eV, 18.8 eV, 16.5 eV, 11.8 eV, 7.05 eV.

Answer from: Quest

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Answer from: Quest

The answer is b.the gaseous state is electrically positive, while the plasma state is electrically negative.

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An electron is trapped in an infinite square-well potential of width 0.4 nm. if the electron is init...

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