Answer :
Answer:
Wavelength, [tex]\lambda=315\ nm[/tex]
Explanation:
It is given that,
Energy in most stable state, [tex]E_1=500\ kJ/mol[/tex]
Energy in its lowest excited state, [tex]E_2=120\ kJ/mol[/tex]
Let [tex]\lambda[/tex] is the wavelength of the radiation emitted when an atom of X undergoes a transition from the lowest excited state to the ground state. It can be calculated as :
[tex]\Delta E=\dfrac{hc}{\lambda}[/tex]
[tex]\lambda=\dfrac{hc}{\Delta E}[/tex]
[tex]\lambda=\dfrac{(6.63\times 10^{-34}J-s)\times (3\times 10^8\ m/s)}{(500-120)\ kJ/mol\times (\dfrac{1}{6.022\times 10^{23}\ mol^{-1}})\times 10^3}[/tex]
[tex]\lambda=3.15\times 10^{-7}\ m[/tex]
or
[tex]\lambda=315\ nm[/tex]
So, the wavelength of the radiation emitted when an atom of X undergoes a transition from the lowest excited state to the ground state is 315 nm. Hence, this is the required solution.