Atoms And Nuclei Question 311
Question: The ionization energy of a hydrogen like Bohr atom is 4 Rydbergs. Find the wavelength of the radiation emitted when the electron jumps from the first excited state to the ground state
Options:
A) $ 300A{}^\circ $
B) $ 2.5\times {{10}^{-11}}m $
C) $ 100A{}^\circ $
D) $ 1.5\times {{10}^{-11}}m $
Show Answer
Answer:
Correct Answer: A
Solution:
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$ E _{n}=-\frac{I.E.}{n^{2}} $ for Bohr “s hydrogen atom. Here, $ I.E/=4R\text{ }\therefore E _{n}=\frac{-4R}{n^{2}} $
$ \therefore E _{2}-E _{1}=\frac{-4R}{2^{2}}-( -\frac{4R}{l^{2}} )=3R $ ..(i) $ E _{2}-E _{1}=hv=\frac{hc}{\lambda } $ .(ii) From (i) and (ii)
$ \therefore \lambda =\frac{hc}{3R}=\frac{6.6\times {{10}^{-34}}\times 3\times 10^{8}}{2.2\times {{10}^{-18}}\times 3}=300\overset{o}{\mathop{A}}, $
