Knockout JEE Main April 2021 (One Month)
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A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :
2
3
4
5
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As we learnt in
Energy emitted due to transition of electron -
- wherein
Energy emitted by Hydrogen atom
So 10.2 eV photon removes an electron from nth orbital of Lithium.
Hence, Energy of nth orbital of Lithium should be - 10.2 eV.
Hence, Value of n = 4
Correct option is 3
Option 1)
2
This is an incorrect option.
Option 2)
3
This is an incorrect option.
Option 3)
4
This is the correct option.
Option 4)
5
This is an incorrect option.