A hydrogen like species (atomic number z) is present in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successive emission of two photons of energies 10.20 ev and 17.0 ev respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successive emission of two photons of energy 4.25 ev and 5.95 ev respectively. Determine the value of z.
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Answer:
Explanation:
As we know,
ΔE = 13.6 Z
2
×(
2
2
1
−
n
2
1
)
For the transition from excited state n to first excited state i.e. n
1
= 2,
(10.20 + 17.00) eV = 13.6 Z
2
×(
2
2
1
−
n
2
1
) ------ equation (1)
For the transition from excited state n to second excited state i.e. n
1
= 3,
(4.24 + 5.95) eV = 13.6 Z
2
×(
3
2
1
−
n
2
1
) ------ equation (2)
On dividing equation 1 and 2,
1.18 =
n
2
−9
n
2
−4
0.18 n
2
= 6.6
n
2
= 36
n = 6
Putting n=6 in equation 1 we get Z=3
Correct answer is D.
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