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Electrons in a certain energy level n =n_(1), can emit 3 spectral lines. When they are in another energy level, n=n_2. They can emit 6 spectral lines. The orbital speed of the electrons in the two orbits are in the ratio |
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Answer» `4:3` `3=(n_(1) (n_(1)-1))/(2)," in first case"`. Or `n_(1)^(2)-n_(1)=6 =0 or (n_(1)-3) (n_(1)+2)=0` Take positive root, `n_(1)=3` Again, `6=(n_(2)(n_(2)-1))/(2)`, in second case. Or `n_(2)^(2)-n_(2)-12=0 or (n_(2)-4) (n_(2)+3)=0` Take positive root, or `n_(2)=4` Now velocity of electron `v=(2PI KZ e^(2))/(nh)` `v_(1)/v_(2)=n_(2)/n_(1)=4/3` |
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