Calculate the de-Broglie wavelength associated with the electron revolving in the first excited state of hydrogen atom. The ground state energy of the hydrogen atom is –13.6 eV.

 

 

Answers (1)
S safeer

\\E_1=\frac{-13.6}{2^2}eV=-5.44\times10^{-19}J

Kinetic energy = - total energy=5.44 x 10-19J

\\\lambda=\frac{h}{\sqrt{2mK}}\\\\=\frac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times5.44\times10^{-19}}}=6.63A^{0}

 

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