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JEE Advanced Model Paper 8 with solutions for online practice
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Question : 10
Total: 84
In a hydrogen like atom, electron is in 2nd excited state, the binding energy of fourth state ofthis atom is 13.6 eV, then
A 25 eV photon can set free the electron from the second excited state of this sample
3 different types of photon will be observed if electrons make transition up to ground statefrom the second excited state
If 23 eV photon is used then K.E. of the ejected electron is 1 eV.
2nd line of balmer series of this sample has same energy value as 1st excitation energy of Hatoms
Validate
Solution:
B.E. of 4th state = 13.6
z
2
n
2
⇒ 13.6
z
2
4
2
= 13.6 ⇒ z = 4
Sample is
B
e
3
+
∴ energy of electron in 3rd state = - 24.17 eV
Therefore 25 eV photon will cause ionization
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