Question

A photon with a wavelength of 2.29 × 10^–7 meter strikes a mercury atom in the ground state.

Answer 1

Detailed solution is attached

Answer 2

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.