Calculate the de-Broglie wavelength of the electron orbitting in the n=2 state of hydrogen atom.

Solution

Given,
n =2
K.E for the second state, E_k, $fraction numerator 13.6 over denominator straight n squared end fraction space equals space fraction numerator 13.6 over denominator 4 end fraction space space space space space space space space space space space equals space 3.4 space x space 1.6 space x space 10 to the power of negative 19 end exponent space J$
De-Broglie wavelength, $straight lambda space equals space fraction numerator straight h over denominator square root of 2 mE subscript straight k end root end fraction$
$equals space fraction numerator 6.63 space straight x space 10 to the power of negative 34 end exponent over denominator square root of 2 space straight x space 9.1 space straight x space 10 to the power of negative 31 end exponent straight x 3.4 straight x 1.6 straight x 10 to the power of negative 19 end exponent end root end fraction equals space 0.067 space straight m$

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Wave Optics

Calculate the de-Broglie wavelength of the electron orbitting in the n=2 state of hydrogen atom.

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