Physics

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.

Determine the energy of this photon in electronvolts.

2 Answer

  •      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.

  • 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            

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