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

Wavelength,
`\lambda=315\ nm`

Step-by-step 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=(hc)/(\lambda)`

`\lambda=(hc)/(\Delta E)`

`\lambda=((6.63* 10^(-34)J-s)* (3* 10^8\ m/s))/((500-120)\ kJ/mol* ((1)/(6.022* 10^(23)\ mol^(-1)))* 10^3)`

`\lambda=3.15* 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.