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
Correct Answer: A
Solution :
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 \[8\sqrt{3}\] \[2\sqrt{3}cm\] \[\sqrt{3}cm\] \[\left( \eta =\text{2}\times \text{1}{{0}^{-\text{4}}}\text{Ns}/{{\text{m}}^{\text{2}}} \right):\] \[6.6\times {{10}^{-6}}N\] \[6.6\times {{10}^{-5}}N\] \[1.32\times {{10}^{-7}}N\] \[13.2\times {{10}^{-7}}N\]
You need to login to perform this action.
You will be redirected in
3 sec
