Answer:
[tex]\lambda=620\ nm[/tex]
[tex]E_p=1.9682\ eV[/tex]
[tex]E_p=206.67\ eV[/tex] is the photon energy of an x-ray
Explanation:
Given:
Energy of photon, [tex]E_p=2\ eV[/tex]
Then according to given:
[tex]E_p=\frac{1240}{\lambda}[/tex]
[tex]2=\frac{1240}{\lambda}[/tex]
[tex]\lambda=620\ nm[/tex] is the wavelength
Again,
Energy of photon, [tex]E_p=?[/tex]
wavelength of the light, [tex]\lambda=630\ nm[/tex]
[tex]E_p=\frac{1240}{630}[/tex]
[tex]E_p=1.9682\ eV[/tex]
Now, photon energy for an x-ray:
wavelength, [tex]\lambda=6\ nm[/tex]
[tex]E_p=\frac{1240}{6}[/tex]
[tex]E_p=206.67\ eV[/tex] is the photon energy of an x-ray