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12.Atoms
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For a certain hypothetical one electron atom the wavelength (in $ \mathring A $ ) for the spectral lines for transition from $n = p$ to $n = 1$ are given by $\lambda = \frac{{1500{p^2}}}{{{p^2} - 1}}$ (where $p > 1$ ), then the ionization potential of this element must be ......$V$ (Take $hc = 12420\ eV- \mathring A $ )
A$0.95$
B$2.05$
C$8.28$
D$13.6$
Solution
${\lambda _{\min }}({\rm{p}} = \infty ){\lambda _{\min }} = 1500\mathop {\rm{A}}\limits^o $
$\Delta \mathrm{E}=-\mathrm{E}_{1}=\frac{12420}{1500} \mathrm{\,eV}=8.28 \mathrm{\,eV}$
Hence ionization potential is $8.28 \mathrm{\,V}$
$\Delta \mathrm{E}=-\mathrm{E}_{1}=\frac{12420}{1500} \mathrm{\,eV}=8.28 \mathrm{\,eV}$
Hence ionization potential is $8.28 \mathrm{\,V}$
Standard 12
Physics
