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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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Consider a hypothitical hydrogen -like atom . The wavelength in A for spectrical lines for transitions from n=P to n=1 are given by $\;\lambda=\large\frac{1500P^2}{P^2-1}\;$ where P=2 , 3 ,4,----. Then wavelength of least energetic and most energetic photons in this series .

$(a)\;2000A^{0}\qquad(b)\;3000A^{0}\qquad(c)\;4000A^{0}\qquad(d)\;5000A^{0}$

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Answer : (a) $\;2000A^{0}$
Explanation :
$\varepsilon=\large\frac{hc}{\lambda}$
For given condition $\;\lambda=\large\frac{1500P^{2} }{P^{2}-1}$
Hence $\;\varepsilon=\large\frac{hc}{\lambda}=\large\frac{hc}{1500}\;(1-\large\frac{1}{P^2})\times10^{10} J$
$=\large\frac{hc}{(1500)\;(1.6\times10^{-19})}\;(1-\large\frac{1}{P^2}) \times10^{10}eV$
$=8.28\;(1-\large\frac{1}{P^2})eV$
Hence energy of $\;n^{th}\;$ state is $\;\varepsilon_{n}=\large\frac{8.28}{n^2}eV$
Maximum energy is released for transition from $\;P=\infty\;$ to $\;P=1\;$ ; hence wavelength of most energetic photon is 1500 $A^{0}\;.$
Least energy is released for transition from $\;n=2\; to \;n=1\;$ transition for P=2 $\;\lambda=2000A^{0}\;.$
answered Feb 26, 2014 by yamini.v
 

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