biophysics 1\nproblem set 1\n1.9 •• neptunium. in the fall of 2002, scientists at los alamos national…

biophysics 1\nproblem set 1\n1.9 •• neptunium. in the fall of 2002, scientists at los alamos national laboratory determined that the critical mass of neptunium - 237 is about 60 kg. the critical mass of a fissionable material is the minimum amount that must be brought together to start a nuclear chain reaction. neptunium - 237 has a density of 19.5 g/cm³. what would be the radius of a sphere of this material that has a critical mass?\n1.10 •• bio bacteria. bacteria vary in size, but a diameter of 2.0 μm is not unusual. what are the volume (in cubic centimeters) and surface area (in square millimeters) of a spherical bacterium of that size?\n1.12 • the volume of a solid cylinder is given by v = πr²h, where r is the radius and h is the height. you measure the radius and height of a thin cylindrical wire and obtain the results r = 0.036 cm and h = 12.1 cm. what do your measurements give for the volume of the wire in mm³? use the correct number of significant figures in your answer.\n1.13 •• a useful and easy - to - remember approximate value for the number of seconds in a year is π×10⁷. determine the percent error in this approximate value. (there are 365.24 days in one year.)

biophysics 1\nproblem set 1\n1.9 •• neptunium. in the fall of 2002, scientists at los alamos national laboratory determined that the critical mass of neptunium - 237 is about 60 kg. the critical mass of a fissionable material is the minimum amount that must be brought together to start a nuclear chain reaction. neptunium - 237 has a density of 19.5 g/cm³. what would be the radius of a sphere of this material that has a critical mass?\n1.10 •• bio bacteria. bacteria vary in size, but a diameter of 2.0 μm is not unusual. what are the volume (in cubic centimeters) and surface area (in square millimeters) of a spherical bacterium of that size?\n1.12 • the volume of a solid cylinder is given by v = πr²h, where r is the radius and h is the height. you measure the radius and height of a thin cylindrical wire and obtain the results r = 0.036 cm and h = 12.1 cm. what do your measurements give for the volume of the wire in mm³? use the correct number of significant figures in your answer.\n1.13 •• a useful and easy - to - remember approximate value for the number of seconds in a year is π×10⁷. determine the percent error in this approximate value. (there are 365.24 days in one year.)

Answer

1.9

Explanation:

Step1: Convert mass to grams

The critical mass is $m = 60\ kg=60\times10^{3}\ g$.

Step2: Use density - mass - volume formula

We know that density $\rho=\frac{m}{V}$, so $V = \frac{m}{\rho}$. Given $\rho = 19.5\ g/cm^{3}$, then $V=\frac{60\times 10^{3}\ g}{19.5\ g/cm^{3}}\approx3076.92\ cm^{3}$.

Step3: Use volume formula for sphere

The volume of a sphere is $V=\frac{4}{3}\pi r^{3}$. We solve for $r$: [r=\left(\frac{3V}{4\pi}\right)^{\frac{1}{3}}=\left(\frac{3\times3076.92\ cm^{3}}{4\pi}\right)^{\frac{1}{3}}] [r=\left(\frac{9230.76}{4\pi}\right)^{\frac{1}{3}}\approx\left(\frac{9230.76}{12.566}\right)^{\frac{1}{3}}\approx(734.56)^{\frac{1}{3}}\approx9.02\ cm]

Answer:

$r\approx9.02\ cm$

1.10

Explanation:

Step1: Convert diameter to radius and units

The diameter of the bacterium $d = 2.0\ \mu m=2.0\times10^{- 4}\ cm$, so the radius $r=\frac{d}{2}=1.0\times10^{-4}\ cm$.

Step2: Calculate volume of sphere

The volume of a sphere $V=\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi(1.0\times 10^{-4}\ cm)^{3}\approx4.19\times10^{-12}\ cm^{3}$.

Step3: Convert radius to millimeters

$r = 1.0\times10^{-4}\ cm = 1.0\times10^{-3}\ mm$.

Step4: Calculate surface - area of sphere

The surface - area of a sphere $A = 4\pi r^{2}=4\pi(1.0\times10^{-3}\ mm)^{2}\approx1.26\times10^{-5}\ mm^{2}$

Answer:

Volume: $V\approx4.19\times 10^{-12}\ cm^{3}$, Surface - area: $A\approx1.26\times10^{-5}\ mm^{2}$

1.12

Explanation:

Step1: Substitute values into volume formula

Given $V=\pi r^{2}h$, $r = 0.036\ cm=0.36\ mm$ and $h = 12.1\ cm = 121\ mm$. [V=\pi(0.36\ mm)^{2}\times121\ mm] [V=\pi\times0.1296\ mm^{2}\times121\ mm] [V\approx49.6\ mm^{3}] (using the correct number of significant figures. Since $r$ has 2 significant figures and $h$ has 3 significant figures, the result should have 2 significant figures)

Answer:

$V\approx49.6\ mm^{3}$

1.13

Explanation:

Step1: Calculate the actual number of seconds in a year

There are $365.24$ days in a year, $24$ hours in a day, $60$ minutes in an hour and $60$ seconds in a minute. So the actual number of seconds $N_{actual}=365.24\times24\times60\times60\ s\approx3.15569\times10^{7}\ s$.

Step2: Calculate the approximate number of seconds

The approximate number of seconds $N_{approx}=\pi\times10^{7}\ s\approx3.14159\times10^{7}\ s$.

Step3: Calculate percent error

The percent - error formula is $\text{Percent Error}=\frac{\vert N_{actual}-N_{approx}\vert}{N_{actual}}\times100%$. [ \text{Percent Error}=\frac{\vert3.15569\times 10^{7}-3.14159\times10^{7}\vert}{3.15569\times10^{7}}\times100%] [=\frac{\vert0.0141\times10^{7}\vert}{3.15569\times10^{7}}\times100%\approx0.45%]

Answer:

$0.45%$