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(a) Using Bohr's postulates, derive the expression for the total energy of the electron in the stationary states of the hydrogen atom. (b) Using Rydberg formula, calculate the wavelengths of the spectral lines of the first member of the Lyman series and of the Balmer series. |
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Answer» Solution :`mvr=(nh)/(2pi)` `(mv^(2))/( r )=(1)/(4PI epsi_(0)).(e)/( r^(2) )` `r=(e^(2))/(4pi epsi_(0) mv^(2))` `r=(Ze^(2))/(4pi epsi_(0)m((nh)/(2pi mr))^(2))` `rArr r=(epsi_(0) n^(2)h^(2))/(pi me^(2))` Potential ENERGY `U=-(1)/(4pi epsi_(0)).(e^(2))/( r )` `(me^(4))/(4 epsi_(0)n^(2)h^(2))` `KE=(1)/(2)mv^(2)` `=(1)/(2)m((nh)/(2pi mr))^(2)` `=(n^(2)h^(2)pi^(2)m^(2)e^(4))/(8pi^(2)me_(0)^(2)n^(4)h^(4))` `KE=(me^(4))/(8epsi_(0)^(2)h^(4))` `KE=(me^(4))/(8epsi_(0)^(2)n^(4)h^(2))` `TE=KE+PE` `=(me^(4))/(8 epsi^(2)n^(2)h^(2))-(me^(4))/(4 sum^(2)jn^(2)h^(2))` `=-(me^(4))/(8 epsi_(0)^(2)n^(2)h^(2))` Rydberg formula : For first MEMBER of Lyman series For first member of Balmer Series `(1)/(lambda)=R((1)/(1^(2))-(1)/(2^(2)))` `=(4)/(3R)` For first member of Balmer Series `(1)/(lambda)=R((1)/(2^(2))-(1)/(3^(2)))` `lambda=(36)/(5R)` |
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