Answered

Calculate the energy of the orange light emitted, per photon, by a neon sign with a frequency of 4.89 × 1014 hz.

Verified Answer :

BladeRunner212

3.24 x 10⁻¹⁹ Joule

Further explanation

Given:

A neon sign with a frequency of 4.89 x 10¹⁴ Hz.

Question:

The energy of the orange light emitted per photon (in joules).

The Process:

The energy of a photon is given by \(\boxed{\boxed{ \ E = hf \ }}\)

  • E = energy in joules
  • h = Planck's constant 6.63 x 10⁻³⁴ Js
  • f = frequency of light in Hz (sometimes the symbol f is written as v)

Let us find out the energy of the orange light emitted per photon.

\(\boxed{ \ E = (6.63 \times 10^{-34})(4.89 \times 10^{14}) \ }\)

Thus, we get a result of \(\boxed{\boxed{ \ E = 3.24 \times 10^{-19} \ J \ }}\)

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Notes:

  • When an electron moves between energy levels it must emit or absorb energy.
  • The energy emitted or absorbed corresponds to the difference between the two allowed energy states, i.e., as packets of light called photons.
  • A higher energy photon corresponds to a higher frequency (shorter wavelength) of light.

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Keywords: calculate, the energy of the orange light emitted, per photon, by a neon sign, frequency, Planck's constant

nuhulawal20

The energy of the orange light emitted, per photon is 3.42 × 10⁻¹⁹J.

Given the data in the question;

  • Frequency; \(f = 4.89 * 10^{14} Hz\)

Energy of the orange light emitted; \(E =\ ?\)

Photon energy is the energy carried by a single photon, represented by the expression:

\(E = hf\)

Where:

  • E is the photon energy,
  • h is the Planck's constants ( \(6.626 * 10^{-34} J Hz^{-1}\) ) and
  • f is the frequency.

We substitute our values into the equitation

\(E = ( 6.626 * 10^{-34}J/Hz) * ( 4.89 * 10^{14} Hz)\\\\E = 3.24 * 10^{-19}J\)

Therefore, the energy of the orange light emitted, per photon is 3.42 × 10⁻¹⁹J

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