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. Determine the energy of this photon
Question
mercury atom falls to a lower energy level.
Determine the energy of this photon in electronvolts.
2 Answer
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1. User Answers Anonym
The energy of an eletromagnetic wave is given by the Planck's Equation:
[tex]E=hf[/tex]
Entering the unknowns:
[tex]E=hf \\ E=6.63*10^{34}*5.48*10^{14} \\ E=36.3324*10^{20}*J[/tex]
Converting unit:
[tex]E= \frac{36.3324*10^{20}*J}{1.602*10^{19}* \frac{J}{eV} } \\ \boxed {E=2.27*10^2eV}[/tex]
Obs: approximate results
If you notice any mistake in my english, please let me know, because i am not native. 
2. User Answers shirleywashington
Answer:
The energy of photon is 2.26 eV
Explanation:
It is given that,
A photon is emitted when an electron in a mercury atom falls to a lower energy level.
Frequency of photon, [tex]f=5.48\times 10^{14}\ Hz[/tex]
We have to find the energy of this photon. Mathematically, the energy of the photon is given by :
E = h × f
Where h is the Planck's constant
[tex]E=6.63\times 10^{34}\times 5.48\times 10^{14}[/tex]
[tex]E=3.63\times 10^{19}\ J[/tex]
We know that : [tex]1\ eV=1.6\times 10^{19}\ J[/tex]
So, [tex]E=\dfrac{3.63\times 10^{19}}{1.6\times 10^{19}}[/tex]
E = 2.26 eV
So, the energy of this photon is 2.26 eV