Asked by DCM
In the hydrogen atom the radius of orbit B is sixteen times greater than the radius of orbit A. The total energy of the electron in orbit A is -0.378 eV. What is the total energy of the electron in orbit B?
Answers
Answered by
Elena
r(B)= 0.53•10^-10•n(B)^2
r(A)= 0.53•10^-10•n(A)^2
r(B)/ r(A) = n(B)^2/ n(A)^2
n(B) = 4•n(A)
E(A) = - 13.6/n(A)^2 eV = - 0.378 eV ,
n(A)^2 = 3.598,
n(A) = 6 ,
n(B) = 4•n(A) = 4•6 = 24.
E(B) = - 13.6/n(B)^2 eV = - 13.6/(24)^2 = - 0.0236 eV.
r(A)= 0.53•10^-10•n(A)^2
r(B)/ r(A) = n(B)^2/ n(A)^2
n(B) = 4•n(A)
E(A) = - 13.6/n(A)^2 eV = - 0.378 eV ,
n(A)^2 = 3.598,
n(A) = 6 ,
n(B) = 4•n(A) = 4•6 = 24.
E(B) = - 13.6/n(B)^2 eV = - 13.6/(24)^2 = - 0.0236 eV.
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