Answered

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.

Verified Answer :

The energy of an eletromagnetic wave is given by the Planck's Equation:

\(E=hf\)
 
     Entering the unknowns:

\(E=hf \\ E=6.63*10^{-34}*5.48*10^{14} \\ E=36.3324*10^{-20}*J\)
 
     Converting unit:

\(E= \frac{36.3324*10^{-20}*J}{1.602*10^{-19}* \frac{J}{eV} } \\ \boxed {E=2.27*10^2eV}\)

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, \(f=5.48\times 10^{14}\ Hz\)

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

\(E=6.63\times 10^{-34}\times 5.48\times 10^{14}\)  

\(E=3.63\times 10^{-19}\ J\)  

We know that :  \(1\ eV=1.6\times 10^{-19}\ J\)

So, \(E=\dfrac{3.63\times 10^{-19}}{1.6\times 10^{-19}}\)

E = 2.26 eV

So, the energy of this photon is 2.26 eV            

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