Saved Bookmarks
| 1. |
Calculatethe energyrequired to excitehydrogenatom fromgroundstateto thesecond excitedstate |
|
Answer» Solution : Datasupplied ` m=9xx 10^(-31)Kg, C = 3 xx 10^8m//s, alpha=(1)/( 137 )` Forgroundstate`n_f= 3` energyrequired` =E_(3!) = E_3- E_1 = (mc ^2 alpha ^2)/(2)[ (1)/(I^2 ) - (1)/( 3 ^2) ] = (mc^2 alpha^2 )/(2)((8)/(9))` ` thereforeE_(31)= ( 9xx 10^(-31)xx(2 xx 10 ^8)^2xx 8) /( 2xx 137 xx 137xx 9)` joules ` E_(31)=( 9xx 9xx 8 xx 10^(-31) xx 10^(16) )/( 2xx9xx 137 xx 137 xx 1.6 xx 10 ^(-19)) eV= 11.98eV ` [ SINCE `, 1 eV= 1.6xx 10^(-19)J`] |
|