Question

The energy of electron in the nth orbit of hydrogen atom is expressed as $ \sqrt{3}cm $ . The shortest and longest wavelength of Lyman series will be :

• A. $ \left( \eta =\text{2}\times \text{1}{{0}^{-\text{4}}}\text{Ns}/{{\text{m}}^{\text{2}}} \right): $
• B. $ 6.6\times {{10}^{-6}}N $
• C. $ 6.6\times {{10}^{-5}}N $
• D. none of these

Answer 1

The Correct Option is A

Energy of electron in nth orbit of hydrogen atom is : $ \text{14 }\times \,\,\text{1}{{0}^{\text{4}}}\text{ K} $ $ \text{1 km}/{{\text{s}}^{\text{2}}} $ $ \text{1}00\text{ m}/{{\text{s}}^{\text{2}}} $ Now for shortest wavelength in Lyman series, the transition of atom takes place from infinity to n = 1 i.e., $ \text{1}0\text{ m}/{{\text{s}}^{\text{2}}} $ $ \text{1}\,\text{m}/{{\text{s}}^{\text{2}}} $ $ \sqrt{98}m/s $ $ \sqrt{\text{49}0}\text{ m}/\text{s} $ $ \sqrt{\text{4}\text{.9}}\text{ m}/\text{s} $ Again for longest wavelength in Lyman series, the transition of electron is from $ 8\sqrt{3} $ $ 2\sqrt{3}cm $ $ \sqrt{3}cm $ $ \left( \eta =\text{2}\times \text{1}{{0}^{-\text{4}}}\text{Ns}/{{\text{m}}^{\text{2}}} \right): $ $ 6.6\times {{10}^{-6}}N $ $ 6.6\times {{10}^{-5}}N $ $ 1.32\times {{10}^{-7}}N $ $ 13.2\times {{10}^{-7}}N $