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Determine the value of thede Broglie wavelength associated with the electron orbiting in the ground state of hydrogen atom ( Given E n = - (13.6)eV and Bohr radius r_(0) = 0.53Å). How will the de Broglie wavelength change when it is the first excited state? |
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Answer» Solution :In ground state, the kinetic energy of the electron is `K = -E = ( + 13.6eV)/( 1^(2)) = 13.6 xx 1.6 xx 10^(-19) J` DE Broglie wavelength, `lambda = ( h )/( p ) = ( h )/( sqrt(2mK))` `lambda_(1)= ( 6.63 xx 10^(-34))/(sqrt( 2 xx 9.1 xx 10^(-31) xx 2.18 xx 10^(-18)))` `= 9.33 xx 10^(-9)= 0.33nm` Kinetic energy in the FIRST EXCITED state ( n = 2 ) `K = - E = + ( 13.6)/( 2^(2)) eV = + 3.4 eV` `= 3.4 xx 1.6 xx 10^(-19) J` `= 0.54 xx 10^(-18) J` de Broglie wavelength, `lambda_(2) = ( h)/( sqrt(2mK))` `= ( 6.63 xx 10^(-34))/( sqrt( 2 xx 9.1 xx 10^(-31) xx 0.544 xx 10^(-15)))` `= 2 xx 0.33 nm =0.66nm` |
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