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
Physics
Anonym
Question
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
units.]
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
units.]
2 Answer

1. User Answers Anonym
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. 
2. User Answers Eduard22sly
The energy of the photon in joules, given that it's has a frequency of 5.48×10¹⁴ Hz is 36.33×10¯²⁰ 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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