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A hydrogen-like atom (atomic number Z) i...

A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.20 eV and 17.00 eV respectively. Alternatively the atom from the same excited state can make a transition to the second excited state by snccessively emitting two photons of energies 4.25 eV and 5.95 eV respectively. Determine the values of n and Z (ionization energy of hydrogen atom = 13.6 eV)

A

n=5

B

z=2

C

n=6

D

z=3

Text Solution

Verified by Experts

The correct Answer is:
C, D

`E= (-2^2)/(n^2) (13.6eV)`
`implies R.hc.z^2 (1/9 - 1/n^2) = 27.2 ….. (1)`
`implies R.hc.z^2 (1/9 - 1/n^2) = 10.2 eV…(2)`
`((1))/((2))= (((n^2 - 4)/(4n^2)))/(((n^2 - 9 )/(9n^2))) = (27.2)/(10.2) = 2.66`
`9(n^2 - 4)xx 4 (n^2 - 9) (2.66)`
`implies 9n^2 - 36 = 10 . 66 n^2 - 96`
`therefore 1.66 n^2 = 60 implies n^2 =(60)/(1.66) = 36`
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