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Calculate the de-Broglie wavelength of the electron orbiting in the n= 2 state of hydrogen atom. |
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Answer» Solution :In n - 2 STATEOF hydrogen atom, as PER Bohr.s quantum condition, we have `mvr_(2) = (nh)/(2pi) = (4h)/(2pi)= (h)/(pi)rArrmv = (h)/(pi r_(2))` `therefore ` de- Broglie wavelenght` lambda = (h)/(MV) = (h)/((h//pi r_(2))) = pi r_(2)` , where `r_(2)`is the radiusof orbitcorrespoding to n = 2 state . |
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