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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Atoms

A doubly ionized lithium atom is hydrogen like atomic number $Z=3$ . Find the wavelength of radiation required to excite electron in $\;Li^{+2}\;$ from first to third Bohr orbit . Given the ionization energy of hydrogen atom as $\;13.6 eV\;.\;$

$(a)\;226.42 A^{0}\qquad(b)\;192.62 A^{0}\qquad(c)\;286.38 A^{0}\qquad(d)\;113.74 A^{0}$

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Answer : (d) $\;113.74 A^{0}$
Explanation :
The energy of $\;n^{th}\;$ orbit of a hydrogen like atom is $\;\varepsilon=-\large\frac{13.6 Z^2}{n^2}$
Thus for $\;Li^{+2}\;$ atom as Z=3 , the electroenergies for first and third Bohr orbits are :
For $\qquad n=1 \;, \varepsilon_{1}=-\large\frac{13.6\times3^2}{1^2} eV$
For $\quad \; n=3 \; , \varepsilon_{3}=-\large\frac{13.6 \times3^2}{3^2}eV$
$=-13.6 eV$
Thus energy required to transformed an electron from $\;\varepsilon_{1}\;$ level to $\;\varepsilon_{3}\;$ level is
$=13.6-(-122.4)=108.8 eV$
The radiation needed to cause this transition should have photons of this energy
$h \nu=108.8 eV$
The wavelength of this radiation is
$\large\frac{hc}{\lambda}=108.8 eV$
or $\quad \lambda=\large\frac{hc}{108.8 eV}=\large\frac{(6.63 \times10^{-34})\times(3 \times 10^{8})}{108.8 \times 1.6\times 10^{-19}} m=113.74 A^{0}\;.$
answered Feb 26, 2014 by yamini.v

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