A) \[102nm\]
B) \[150nm\]
C) \[82nm\]
D) \[122nm\]
Correct Answer: D
Solution :
: For any series, the transition that produces the least energetic photon is the transition between the home-base level that defines the series and the level immediately above it. For the Lyman series, the home-base level is at \[n=1\]. So the transition that produces the least energetic photon is the transition from the \[n=2\]level to the\[n=1\] level. \[\therefore \]The wavelength for the least energetic photon is \[\lambda =\frac{hc}{{{E}_{2}}-{{E}_{1}}}\] Here, \[he=1240\text{ }eVnm\] \[{{E}_{1}}=-\frac{13.6}{{{1}^{2}}}eV=-13.6eV\] (as \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV\]) \[{{E}_{2}}=-\frac{13.6}{{{2}^{2}}}eV=-3.4eV\] \[\therefore \] \[\lambda =\frac{1240\,eV\,nm}{-3.4eV-(-13.6eV)}\] \[=\frac{1240\,eV\,nm}{10.2\,eV}=122nm\]
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