Atoms And Nuclei Question 299
Question: The ionization energy of a hydrogen-like Bohr atom is 4 Rydbergs. Find the wavelength of radiation emitted when the electron jumps from the first excited state to the ground state: [1 Rydberg $ =2.2\times {{10}^{-18}},h=6.6\times {{10}^{-34}}Js, $ $ c=3\times 10^{8}m/s $ . Bohr radius of hydrogen atom $ =5\times {{10}^{-11}}m $
Options:
A) $ 400\overset{o}{\mathop{A}}, $
B) $ 300\overset{o}{\mathop{A}}, $
C) $ 500\overset{o}{\mathop{A}}, $
D) $ 600\overset{o}{\mathop{A}}, $
Show Answer
Answer:
Correct Answer: B
Solution:
- The energy in the ground state $ E_{1}=-,4Rydberg\text{ }E_{1}=-,4\times 2.2\times {{10}^{-18}}J $ The energy of the first excited state (n=2) $ E_{2}=\frac{E_{1}}{4}=2.2\times {{10}^{-18}}J $ The energy difference $ \Delta E=E_{2}-E_{1}=3\times 2.2\times {{10}^{-18}}J $ Now, the wavelength of radiation emitted is $ \lambda =\frac{hc}{\Delta E} $ $ \lambda =\frac{6.6\times {{10}^{-34}}\times 3\times 10^{8}}{3\times 2.2\times {{10}^{-18}}}=300\overset{o}{\mathop{A}}, $
