Question

This problem has been solved!See the answerFrom the information in this chapter on the mass of the proton, the mass of the electron, and the sizes of the nucleus and the atom, calculate the densities of a hydrogen nucleus and a hydrogen atom. (Radius of hydrogen atom is 1 ✕ 10-8 cm.)Can someone please help with this two part question? I have asked twice on here and I am still getting it wrong on my WebAssign-it's the only one I cannot get!.... Thank you in advance!!density of hydrogen nucleus :

Answer 1

The density of a hydrogen nucleus is approximately 1.24 x 10^17 kg/m^3, and the density of a hydrogen atom is approximately 8.38 x 10^16 kg/m^3.

Step-by-step explanation:

To calculate the density of a hydrogen nucleus, we can use the formula for density, which is mass divided by volume. The mass of a proton is approximately 1.67 x 10^-27 kg, and the volume of a sphere can be calculated using the formula V = (4/3)πr^3. Let's assume the radius of a hydrogen nucleus is equal to the diameter of a proton, which is approximately 1.2 fm. Plugging these values into the formula, we can calculate the density of a hydrogen nucleus.

density = mass / volume = 1.67 x 10^-27 kg / ((4/3)π(0.6 x 10^-15 m)^3) = 1.24 x 10^17 kg/m^3

So, the density of a hydrogen nucleus is approximately 1.24 x 10^17 kg/m^3.

To calculate the density of a hydrogen atom, we need to know its mass and volume. The mass of an electron is approximately 9.11 x 10^-31 kg, and the volume of a sphere can be calculated using the formula V = (4/3)πr^3. Let's assume the radius of a hydrogen atom is equal to the given value, which is 1 x 10^-8 cm. Plugging these values into the formula, we can calculate the density of a hydrogen atom.

density = mass / volume = 9.11 x 10^-31 kg / ((4/3)π(0.5 x 10^-10 m)^3) = 8.38 x 10^16 kg/m^3

So, the density of a hydrogen atom is approximately 8.38 x 10^16 kg/m^3.

Answer 2

The density of the hydrogen atom is 0,4 g/cm3 and the density of the hydrogen nucleus is 7,8x10^14 g/cm3.

Step-by-step explanation:

Hydrogen is the first element of the periodic table
`H^(1) _(1)`.

The atomic number of an element tell us the number of protons (Z), and the atomic mass (A) tell us the number of protons plus neutrons. The number of electrons is always the same as protons. Therefore, for the hydrogen:

(Z): 1=# protons

(A): 1=#protons + #neutrons

Solving this system, we have for hydrogen:

#protons=1

#electrons=1

#neutrons=0

Density (d) is defined as the relation between mass and volumen:
`d=(m)/(V)`.

Density of the hydrogen nucleus:

In the nucleus we have the protons and neutrons, as the hydrogen doesn't have neutrons, we onlye have one proton in its nucleus. The mass of the proton is 1,67262 x
`10^(-24)` g, therefore the mass of the nucleus is 1,67262x
`10^(-24)` g.

Will find the volume of the nucleus assuming it has a spherical shape. The radius of the hydrogen nucleous is 8 x
`10^(-14)` cm. To calculate the volume of an sphere we use the following equation:

v =
`(4)/(3)\pi*r^(3)`

`v=(4)/(3)*\pi*(8*10^(-14) )^(3) = 2,144*10^(-39) cm^(3)`

Finally, to calculate the density we only replace the mass and volumen in the density equation:

`d=(1,67262*10^(-24) )/(2,144*10^(-39) )= 7,8*10^(14) (g)/(cm^(3))`

Density of the hydrogen atom:

In the atom we have all the electrons, neutrons and protons. For the hydrogen atom we have 1 proton and 1 electron. The mass of one electron is 9,1x
`10^(-28)` g. So, the mass of the hydrogen atom is 9,1x
`10^(-28)` kg + 1,67262x
`10^(-24)` = 1,67353x
`10^(-24)</u>` g.

To find the volume we use the equation of the sphere, but now using the value of 1x
`10^(-8)` cm as radius.

`v=(4)/(y)\pi(10^(-8) )^(3) = 4,189*10^(-24) cm3`

And the density will be:

`d = (1,67353*10^(-24) )/(4,189*10^(-24) )= 0,4 (g)/(cm^(3) )`