Question

When an electron in an atom absorbed a photon, the energy of the electron changed by 2.94 x 10-19 J. What is the frequency of the photon in Hertz (1/s)

Answer 1

When an electron in an atom absorbed a photon, the energy of the electron changed by 2.94 × 10⁻¹⁹. The frequency of the photon is 4.4 × 10¹⁴ Hertz .

How to calculate the frequency of photon ?

To calculate the frequency of photon use the expression

E = hv

where,

E = Energy

h = Planck's Constant

v = frequency (Hz)

Put the value in above formula we get

E = hv

2.94 × 10⁻¹⁹ = 6.6 × 10⁻³⁴ × v

v = $\frac{2.94 \times 10^{-19}\ J}{6.6 \times 10^{-34}}$

  = 0.44 × 10¹⁵

  = 4.4 × 10¹⁴

Thus from the above conclusion we can say that When an electron in an atom absorbed a photon, the energy of the electron changed by 2.94 × 10⁻¹⁹. The frequency of the photon is 4.4 × 10¹⁴ Hertz .

Learn more about the Plank's Constant here: brainly.com/question/27163891

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