A) 2
B) 4
C) 5
D) 6
Correct Answer: B
Solution :
Energy of radiations\[E=\frac{hc}{\lambda }\] \[\begin{align} & =\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{975\times {{10}^{-10}}\times 1.6\times {{10}^{-19}}}eV \\ & =12.75\,eV \\ \end{align}\] Energy of hydrogen atom in first energy level =-13.6 eV Energy of atom after absorbing radiations =-13.6+12.75=-0.85 eV Number of spectral lines \[\begin{align} E=\frac{13.6}{{{n}^{2}}} & \\ {{n}^{2}}=\frac{13.6}{0.85} & \\ \end{align}\] \[{{n}^{2}}=16\Rightarrow n=4\]You need to login to perform this action.
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