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Find the wavelength of electron orbiting in the first excited state of hydrogen atom. |
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Answer» Here, `lambda`=? n=2 for first excited state `v=(2piKe^2)/(nh)` `lambda=h/(mv)=(h,nh)/(2pimKe^2)=(nh^2)/(2pimKe^2)` `lambda=(2(6.6xx10^(-34))^(2))/(2xx3.14xx9xx10^(-31)(9xx10^9)(1.6xx10^(-19))^(2))` `lambda=(2xx6.6xx6.6xx10^-8m)/(6.28xx81xx1.6xx1.6)=6.69xx10^(-10)m` |
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