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Chemistry Question on Quantum Mechanical Model of Atom

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

A

(η=2×10−4Ns/m2):\left( \eta =\text{2}\times \text{1}{{0}^{-\text{4}}}\text{Ns}/{{\text{m}}^{\text{2}}} \right):

B

6.6×10−6N6.6\times {{10}^{-6}}N

C

6.6×10−5N6.6\times {{10}^{-5}}N

D

none of these

Answer

(η=2×10−4Ns/m2):\left( \eta =\text{2}\times \text{1}{{0}^{-\text{4}}}\text{Ns}/{{\text{m}}^{\text{2}}} \right):

Explanation

Solution

Energy of electron in nth orbit of hydrogen atom is : 14 ×  104 K\text{14 }\times \,\,\text{1}{{0}^{\text{4}}}\text{ K} 1 km/s2\text{1 km}/{{\text{s}}^{\text{2}}} 100 m/s2\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., 10 m/s2\text{1}0\text{ m}/{{\text{s}}^{\text{2}}} 1 m/s2\text{1}\,\text{m}/{{\text{s}}^{\text{2}}} 98m/s\sqrt{98}m/s 490 m/s\sqrt{\text{49}0}\text{ m}/\text{s} 4.9 m/s\sqrt{\text{4}\text{.9}}\text{ m}/\text{s} Again for longest wavelength in Lyman series, the transition of electron is from 838\sqrt{3} 23cm2\sqrt{3}cm 3cm\sqrt{3}cm (η=2×10−4Ns/m2):\left( \eta =\text{2}\times \text{1}{{0}^{-\text{4}}}\text{Ns}/{{\text{m}}^{\text{2}}} \right): 6.6×10−6N6.6\times {{10}^{-6}}N 6.6×10−5N6.6\times {{10}^{-5}}N 1.32×10−7N1.32\times {{10}^{-7}}N 13.2×10−7N13.2\times {{10}^{-7}}N