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.

Answer 1

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= (36.3324*10^(-20)*J)/(1.602*10^(-19)* (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 2

The energy of photon is 2.26 eV

Step-by-step 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* 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* 10^(-34)* 5.48* 10^(14)`

`E=3.63* 10^(-19)\ J`

We know that :
`1\ eV=1.6* 10^(-19)\ J`

So,
`E=(3.63* 10^(-19))/(1.6* 10^(-19))`

E = 2.26 eV

So, the energy of this photon is 2.26 eV