The nucleus of a 15 a distance of 5.4 x 10-11 m. What is the ratio of the density of the hydrogen nucleus to the density of the complete hydrogen atom? Section 30.1 10 30.P energy as a ground-state For a doubly ionized lithium atom Li2+ (z - 3), what is the principal quantum number of the state in which the electron has the same total Section 30.3 CJ10 36 P010 Solution 2) First find density of proton mass of proton = 1.67e-27 V = 4/3*pi*r 3 = 4/3*pi*(1.7e-15) 3 = 2.0579e-44 m 3 density = mass / volume = 1.67e-27 / 2.0579e-44 density = 8.115e16 kg/m 3 Mass of hydrogen atom = 1.67e-27 V = 4/2*pi*(5.4e-11) 3 = 6.596e-31 m 3 density = 1.67e-27 / 6.596e-31 density = 2531.9 kg/m3 Ratio =  8.115e16 / 2531.9 Ratio = 3.205e13 3) n 2 (Li) = (n 2 / Z 2 ) Hydrogen * Z 2 (Li) for ground state, n and z = 1 for hydrogen n 2 = 1*3 2 = 9 n (Li) = 3 .

1. The nucleus of a 15 a distance of 5.4 x 10-11 m. What is the ratio of the density of the hydrogen
nucleus to the density of the complete hydrogen atom? Section 30.1 10 30.P energy as a ground-
state For a doubly ionized lithium atom Li2+ (z - 3), what is the principal quantum number of the
state in which the electron has the same total Section 30.3 CJ10 36 P010
Solution
2) First find density of proton
mass of proton = 1.67e-27
V = 4/3*pi*r 3
= 4/3*pi*(1.7e-15) 3
= 2.0579e-44 m 3
density = mass / volume = 1.67e-27 / 2.0579e-44
density = 8.115e16 kg/m 3
Mass of hydrogen atom = 1.67e-27
V = 4/2*pi*(5.4e-11) 3
= 6.596e-31 m 3
density = 1.67e-27 / 6.596e-31
density = 2531.9 kg/m3
Ratio =Â Â 8.115e16 / 2531.9
Ratio = 3.205e13
3) n 2
(Li) = (n 2
/ Z 2
) Hydrogen * Z 2
(Li)
for ground state, n and z = 1 for hydrogen
n 2
= 1*3 2
= 9
n (Li) = 3