Some of the energy states of a hypothetical atom, in units of electron volts, ar

ID: 1498950 • Letter: S

Question

Some of the energy states of a hypothetical atom, in units of electron volts, are E1 = -31.50, E2 = -13.10, E3 = -5.20, E4 = -3.60.

Part A

Determine the energy of the least energetic photon that can be absorbed by these atoms when initially in their ground state.

Express your answer to three significant figures and include the appropriate units.

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Correct

Part B

Determine the wavelength of the least energetic photon that can be absorbed by these atoms when initially in their ground state.

Express your answer to three significant figures and include the appropriate units.

Some of the energy states of a hypothetical atom, in units of electron volts, are E1 = -31.50, E2 = -13.10, E3 = -5.20, E4 = -3.60.

Part A

Determine the energy of the least energetic photon that can be absorbed by these atoms when initially in their ground state.

Express your answer to three significant figures and include the appropriate units.

E = 18.4 eV

SubmitMy AnswersGive Up

Correct

Part B

Determine the wavelength of the least energetic photon that can be absorbed by these atoms when initially in their ground state.

Express your answer to three significant figures and include the appropriate units.

given that

E1 = -31.50 eV ( energy at ground state)

E2 = -13.10 eV ( energy at first excited stage)

E3 = -5.20 eV

E4 = -3.60 eV

energy of the least energetic photon = E2 - E1

E = -13.10 -(-31.50)

E = 18.40 eV

(b)

we know that

E = h*c / lambda

where h = planck's constant = 6.6*10(-34) Js

c = speed of light = 3*108 m/s

lambda = h*c / E

lambda = 6.6*10(-34) Js * 6.24*10(18) eV * 3*108 m/s / 18.40 eV

lambda = 0.671*10(-7) m = 67.1 nm

wavelength of the least energetic photon = 67.1 nm

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