Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission
spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than
the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV.
The value of Z is __________.
Answers
Answered by
0
Answer:
ANSWER
△E
2−1
=13.6×z
2
[1−
4
1
]=13.6×z
2
[
4
3
]
△E
3−2
=13.6×z
2
[
4
1
−
9
1
]=13.6×z
2
[
36
5
]
△E
2−1
=△E
3−2
+74.8
13.6×z
2
[
4
3
]=13.6×z
2
[
36
5
]+74.8
13.6×z
2
[
4
3
−
36
5
]=74.8
z
2
=9
z=+3 (z can't be - ve)
Similar questions