Answer:
Energy of photon, \(E=8.64\times 10^{-19}\ J\)
Explanation:
It is given that,
Wavelength of photon, \(\lambda=2.29\times 10^{-7}\ m\)
It strikes a mercury atom in the ground state. We have to find the energy of this photon. It can be calculated using below relation as :
\(E=\dfrac{hc}{\lambda}\)
Where
h is the Planck's constant
c is the speed of light
So, \(E=\dfrac{6.6\times 10^{-34}\ J-s\times 3\times 10^8\ m/s}{2.29\times 10^{-7}\ m}\)
\(E=8.64\times 10^{-19}\ J\)
Hence, the above value is the energy of this photon.
