Question

What is the wavelength of an X-ray photon with energy 8.0 keV (8000 eV)? (1 eV = 1.60 × 10−19 joule.)

What is its frequency?

Answer 1

The wavelength of the X-ray photon is 1.55 nm, and its frequency is 1.93 x 10^17 Hz.

Step-by-step explanation:

To find the wavelength of an X-ray photon with an energy of 8.0 keV, we can use the equation:

λ = c / f

where λ is the wavelength, c is the speed of light (3.00 x 10^8 m/s), and f is the frequency.

First, we need to convert the energy to Joules:

E = 8.0 keV * (1.60 x 10^-19 J/eV) = 1.28 x 10^-16 J

Next, we can use the equation:

E = hf

where h is Planck's constant (6.63 x 10^-34 J·s) and f is the frequency. Rearranging the equation, we get:

f = E / h = (1.28 x 10^-16 J) / (6.63 x 10^-34 J·s) = 1.93 x 10^17 Hz

Now we can substitute the values into the wavelength equation:

λ = (3.00 x 10^8 m/s) / (1.93 x 10^17 Hz) = 1.55 x 10^-9 m, or 1.55 nm.

Therefore, the wavelength of the X-ray photon is 1.55 nm, and its frequency is 1.93 x 10^17 Hz.

Answer 2

Wavelength = 0.15 nm

Frequency =
`1939.3939* 10^(15)Hz`

Step-by-step explanation:

We have given photon energy E = 8 KeV = 8000 eV

In question it is given that
`1eV=1.6* 10^(-19)J`

So
`8000eV=1.6* 8000* 10^(-19)=12800* 10^(-19)j`

Plank's constant
`h=6.6* 10^(-34)js`

We know that photon energy is given by
`E=h\\u`

So
`12800* 10^(-19)=6.6* 10^(-34)\\u`

`\\u =1939.3939* 10^(15)Hz`

Now wavelength
`\lambda =(c)/(f)=(3* 10^8)/(1939.3939* 10^(15))=0.0015* 10^(-7)m=0.15nm`