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An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let lamda_(n),lamda_(g) be the de Broglie wavelength of the electron in the n^(th) state and the ground state respectively. Let A_(n) be the wavelength of the emitted photon in the transition from the n^(th) state to the ground state. For large n, (A, B are constant) |
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Answer» `A_(n)~~A+(B)/(lamda_(n)^(2))` but `K_(n)=(h^(2))/(2mlamdan^(2))` and `K_(g)=(h^(2))/(2mlamdag^(2))` `:.K_(g)-K_(n)=(h^(2))/(2m)[(1)/(lamda_(g)^(2))-(1)/(lamda_(n)^(2))]` but `K_(n)=-E_(n)` for emission of PHOTON and `K_(g)-E_(g)` `:.E_(n)-E_(g)=(HC)/(lamda_(n))` `K_(g)-K_(n)=(hc)/(A_(n))` `A_(n)=(hc)/(k_(g)-k_(n))=(hc)/((h^(2))/(2m)[(1)/(lamda_(g)^(2))-(1)/(lamda_(n)^(2))])` `A_(n)=(2mc)/(h[(lamda_(n)^(2)-lamda_(g)^(2))/(lamda_(n)^(2)lamda_(g)^(2))])=(2mclamda_(n)^(2)lamda_(g)^(2))/(h(lamda_(n)^(2)-lamda_(g)^(2)))` `A_(n)=(2mclamda_(g)^(2))/(h[1-(lamda_(g)^(2))/(lamda_(n)^(2))])=(2mclamda_(g)^(2))/(h)[1-(lamda_(g)^(2))/(lamda_(n)^(2))]^(-1)` `A_(n)=(2mclamda_(g)^(2))/(h)[1+(lamda_(g)^(2))/(lamda_(n)^(2))]` `A_(n)=(2mclamda_(g)^(2))/(h)+(2mclamda_(g)^(4))/(hlamda_(n)^(2))` `=A+(B)/(lamda_(n)^(2))` where `A=(2mclamda_(g)^(2))/(h)` and `B=(2mclamda_(g)^(4))/(h)` |
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