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A hydrogen like atom (atomic No. Z) is in a higher excited state of quantam 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 successively emitting two photons of energy 4.25eV and 5.95 eV respectively. Determine the values of n and Z.

$(a)\; n=6,z = 3\qquad(b)\;n=3,z=6\qquad(c)\;n=2,z=5\qquad(d)\;n=4,n=2$

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Total energy liberated during transition of electron from $n^{th}$ shell to first excited state
(i.e $2^{nd}$ shell) = 10.20+17.0 = 27.20 eV
= $27.20\times1.602\times10^{-12}$ erg
$\large\frac{hc}{\lambda}$= $R_H \times Z^2hc[\large\frac{1}{2^2}-\large\frac{1}{n^2}]$
$27.20\times1.602\times10^{-12}$ = $R_H \times Z^2hc[\large\frac{1}{2^2}-\large\frac{1}{n^2}]$..... (1)
Similarly total energy liberated during transition of electrons from $n^{th}$ shell to second excited state
(i.e $3^{rd}$ shell ) = 4.25+5.95 = 10.20 eV
= $10.20\times1.602\times10^{-12}$ erg
$\therefore10.20\times1.602\times10^{-12}$ = $R_H \times Z^2hc[\large\frac{1}{3^2}-\large\frac{1}{n^2}]$..... (2)
Dividing Eq (1) by Eq (2)
We get n = 6
On substituting the value of n in equation (1) or (2) we get z=3
Hence answer is (a)
answered Jan 28, 2014 by sharmaaparna1
 

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