Question

The energy needed to ionize an atom of element X when it is in its most stable state is 500 kJ mol21. However, if an atom of X is in its lowest excited state, only 120 kJ mol21 is needed to ionize it. What is the wavelength of the radia- tion emitted when an atom of X undergoes a transition from the lowest excited state to the ground state?

Answer 1

Answer:

Wavelength, $\lambda=315\ nm$

Explanation:

It is given that,

Energy in most stable state, $E_1=500\ kJ/mol$

Energy in its lowest excited  state, $E_2=120\ kJ/mol$

Let $\lambda$ 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 :

$\Delta E=\dfrac{hc}{\lambda}$

$\lambda=\dfrac{hc}{\Delta E}$  

$\lambda=\dfrac{(6.63\times 10^{-34}J-s)\times (3\times 10^8\ m/s)}{(500-120)\ kJ/mol\times (\dfrac{1}{6.022\times 10^{23}\ mol^{-1}})\times 10^3}$

$\lambda=3.15\times 10^{-7}\ m$

or

$\lambda=315\ nm$

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.