Answer:
Potential energy for n = 6 Bohr orbit electron is -1.21*10⁻¹⁹J
Explanation:
As per the Bohr model, the potential energy of electron in an nth orbit is given as:
[tex]PE_{n} = -\frac{kZe^{2}}{r_{n}}[/tex]
here:
k = Coulomb's constant = 9*10⁹ Nm2/C2
Z = nuclear charge
e = electron charge = 1.6*10⁻¹⁹ C
r(n) = radius of the nth orbit = n²(5.29*10⁻¹¹m)
Substituting for k, Z(= 1), e and r(n) in the above equation gives:
[tex]PE_{6} = -\frac{9*10^{9}Nm2/C2*1*(1.6*10^{-19}C)^{2}}{(6)^{2}*5.29*10^{-11}m}=-1.21*10^{-19}J[/tex]
