A photon with a frequency of 5.48 × 10^14 hertz is emitted when an electron in a
mercury atom falls to a lower energy level.

Calculate the energy of this photon in joules. [Show all work, including the equation and substitution with

2 Answer

  •      Using the Planck's Equation, comes:

    [tex]E=hf \\ E=6.63*10^{-34}*5.48*10^{14}*J \\ \boxed {E=36.3324*10^{-20}*J}[/tex]

    If you notice any mistake in my english, please let me know, because i am not native.
  • The energy of the photon in joules, given that it's has a frequency of 5.48×10¹⁴ Hz is 36.33×10¯² J

    Data obtained from the question

    • Frequency (f) = 5.48×10¹⁴ Hz
    • Energy (E) =?

    How to determine the energy

    • Planck's constant (h) = 6.63×10¯³⁴ Js
    • Frequency (f) = 5.48×10¹⁴ Hz
    • Energy (E) =?

    The energy of the photon can be obtained as follow:

    E = hf

    E = 6.63×10¯³⁴ × 5.48×10¹⁴

    E = 36.33×10¯²⁰ J

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