Menu div.question,div.steps {font-size: 16px; padding: 10px;line-height: 20px; margin: 20px;} div.question{border: 1px solid #ccc;} div#solution .button{font-size: 20px; margin: 20px; width: 200px; padding: 10px; text-align: center;} div.hint,div.answer,div.steps{display: none;} div.pre_next{font-size: 30px; width:90%; margin:auto;} div.pre_next a.nav_pre{float:left;} div.pre_next a.nav_next{float:right;} div.title div.notes{ font-family:sans-serif, Arial; font-size: 18px; } 2. A proton, which is the nucleus of a hydrogen atom, can be modeled as a sphere with a diameter of 2.4 fm and a mass of 1.67 \times 10^(-27) kg. (a)Determine the density of the proton. (b) State how your answer to part (a) compares with the density of osmium, given in Table 14.1 Hint 1fm = 10^(-15) m, ρ=m/V, V=4/3πr^3 Answer (a) ρ_p = 2.3 times 10^(17) kg//m^3 (b) The density of a proton is 1.02 times 10^(13)  times of the density of osmium. Show Steps Knowns: Diameter of proton: d_p=2.4fm => r_p = 1.2fm = 1.2 times 10(-15) m Mass of proton: m_p = 1.67 times 10^(-27) kg (a) Density of proton: ρ_p = m_p /V_p = m_p/(4/3πr_p^3) ρ_p = (1.67 times 10^(_27) kg) /(4/3π(1.2 times 10^(-15))^3) ρ_p = 2.3times 10^(17) kg//m^3 (b) Let's pick osmium as a sample in table 14.1 ρ_(os) = 22.6 times 10^3 kg//m^3 ρ_p/ρ_(os) = (2.3times10^(17) kg//m^3)/(22.6 times 10^3 kg//m^3) = 1.02 times 10^(13) So the density of a proton is 1.02 times 10^(13)  times of the density of osmium.