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?
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.
To find the total energy of the electron in orbit B, we can use the formula for the energy of an electron in the hydrogen atom:
E = -13.6 eV / n^2
Where E is the total energy, n is the principal quantum number, and -13.6 eV is the ionization energy of hydrogen.
From the given information, we know that the radius of orbit B is 16 times greater than the radius of orbit A.
The radius of an electron orbit is directly proportional to the square of the principal quantum number (r ∝ n^2).
So, we can say that the principal quantum number of orbit B (n_B) is equal to the square root of 16 times the principal quantum number of orbit A (n_A).
n_B = √(16 * n_A)
Since the total energy depends on the square of the principal quantum number, we can substitute the values for n_B and n_A in the energy formula to find the total energy in orbit B.
E_B = -13.6 eV / (n_B)^2
Substituting n_B = √(16 * n_A) into the formula:
E_B = -13.6 eV / (√(16 * n_A))^2
Simplifying:
E_B = -13.6 eV / 16 * n_A
The total energy of the electron in orbit A is given as -0.378 eV. Substituting this value into the equation, we get:
E_B = -13.6 eV / 16 * (-0.378 eV)
Simplifying:
E_B = -13.6 eV / -6.048 eV
E_B ≈ 2.25 eV
Therefore, the total energy of the electron in orbit B is approximately 2.25 eV.