1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Natalka [10]
3 years ago
10

A 70 kg human body typically contains 140 g of potassium. Potassium has a chemical atomic mass of 39.1 u and has three naturally

occurring isotopes. One of those isotopes, 40K (potassium), is radioactive with a half-life of 1.3 billion years and a natural abundance of 0.012 %. Each 40K (potassium) decay deposits, on average, 1.0 MeV of energy into the body.
What yearly dose in Gy does the typical person receive from the decay of 40K (potassium) in the body? Express your answer using two significant figures.
Physics
1 answer:
KatRina [158]3 years ago
8 0

Answer:

Gy = 3.14x10⁻⁴ Gy

Explanation:

To get the dose in Gy we need to use the following expression:

Gy = E / m  (1)

Where:

Gy: dose

E: energy absorbed per atom

m: mass of the human body.

We don't have the energy per atom, but we can calculate that by following the next procedure.

First, let's determine the number of atoms of potassium in our body. For that we need to determine the moles in the 140 g of potassium, with the molecular mass and then, use the avogadro's number:

moles = m/MM

moles = 140 / 39.1 = 3.58 moles

N° atoms = 3.58 * 6.02x10²³ atoms = 2.16x10²⁴ atoms of K.

The abundance of the ⁴⁰K is 0.012% so the atoms of this isotope would be:

N = 2.16x10²⁴ * (0.012/100) = 2.59x10²⁰ atoms of ⁴⁰K.

With this number, and the half life rate, we can determine the number of decay atoms in a year (λ) using the following expression:

λ = ln2 / t(1/2)

λ = ln2 / 1.3x10⁹ = 5.33x10⁻¹⁰ year⁻¹

This number, multiplied by the number of atoms:

R = 5.33x10⁻¹⁰ * 2.59x10²⁰ = 1.38x10¹¹ atoms/year

Now, each atom of K gives an average energy of 1 MeV, so with the atoms we have:

E = 1.38x10¹¹ * 1x10⁶ eV = 1.38x10¹⁷ eV

This value can be expressed in Joules so:

E = 1.38x10¹⁷ eV * (1 J / 6.24x10¹⁸ eV) = 0.022 J

Finally, we can use (1) to get the dose in Gy:

Gy = 0.022 / 70

<h2>Gy = 3.14x10⁻⁴ Gy</h2><h2></h2>

Hope this helps

You might be interested in
The planet Uranus has a radius of 25,360 km and a surface acceleration due to gravity of 9.0 m/s^2 at its poles. Its moon Mirand
AlexFokin [52]

Answer:

8.67791\times 10^{25}\ kg

0.34589\ m/s^2

0.07903\ m/s^2

Explanation:

M = Mass of Uranus

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Uranus = 25360 km

h = Altitude = 104000 km

r_m = Radius of Miranda = 236 km

m = Mass of Miranda = 6.6\times 10^{19}\ kg

Acceleration due to gravity is given by

g=\dfrac{GM}{r^2}\\\Rightarrow M=\dfrac{gr^2}{G}\\\Rightarrow M=\dfrac{9\times 25360000^2}{6.67\times 10^{-11}}\\\Rightarrow M=8.67791\times 10^{25}\ kg

The mass of Uranus is 8.67791\times 10^{25}\ kg

Acceleration is given by

a_m=\dfrac{GM}{(r+h)^2}\\\Rightarrow a_m=\dfrac{6.67\times 10^{-11}\times 8.67791\times 10^{25}}{(25360000+104000000)^2}\\\Rightarrow a_m=0.34589\ m/s^2

Miranda's acceleration due to its orbital motion about Uranus is 0.34589\ m/s^2

On Miranda

g_m=\dfrac{Gm}{r_m^2}\\\Rightarrow g_m=\dfrac{6.67\times 10^{-11}\times 6.6\times 10^{19}}{236000^2}\\\Rightarrow g_m=0.07903\ m/s^2

Acceleration due to Miranda's gravity at the surface of Miranda is 0.07903\ m/s^2

No, both the objects will fall towards Uranus. Also, they are not stationary.

6 0
3 years ago
Neo and the agent stand in different locations and the length of the line between them (their distance apart) is a measurement o
scZoUnD [109]
I think their distance is a measurement of : B. space in two dimension

In two-dimensional space, both directions located in the same plane , and the distance in locations only separated by width and length (there is no volume in this model)
5 0
3 years ago
Read 2 more answers
13 points!
skelet666 [1.2K]
I think it may be C!
4 0
3 years ago
Read 2 more answers
Consider a stone in free fall on a planet with gravitational acceleration 3.4 m/s^2. Suppose you would like the stone to experie
Stella [2.4K]

Answer:

Angle of incline is 20.2978°

Explanation:

Given that;

Gravitational acceleration on a planet a = 3.4 m/s²

Gravitational acceleration on Earth g = 9.8 m/s²

Angle of incline = ∅

Mass of the stone = m

Force on the stone along the incline will be;

F = mgSin∅

F = ma

The stone has the same acceleration as that of the gravitational acceleration on the planet.

so

ma = mgSin∅

a = gSin∅

Sin∅ = a / g

we substitute

Sin∅ = (3.4 m/s²) / (9.8 m/s²)

Sin∅ = 0.3469

∅ = Sin⁻¹( 0.3469 )

∅ = 20.2978°

Therefore, Angle of incline is 20.2978°

8 0
3 years ago
Where does the majority of the mass of an atom come from
asambeis [7]

Answer:

The Nucleus

Explanation:

The nucleus contains the majority of an atom's mass because protons and neutrons are much heavier than electrons, whereas electrons occupy almost all of an atom's volume.  I hope this helps you :D

5 0
3 years ago
Read 2 more answers
Other questions:
  • What is the difference between speed and velocity? A. Velocity takes acceleration into account, while speed does not. B. Speed t
    9·1 answer
  • How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the o
    8·1 answer
  • What is the significance of the direction of an electric field line at some point on the line?
    8·1 answer
  • A child of mass 49.7 kg sits on the edge of a merry-go-round with radius 2.7 m and moment of inertia 311.589 kg m2 . The merrygo
    6·1 answer
  • One kind of baseball pitching machine works by rotating a light and stiff rigid rod about a horizontal axis until the ball is mo
    13·1 answer
  • What came before the big bang? What was..there? If not matter?
    11·1 answer
  • URGENT
    11·1 answer
  • The unit for acceleration is always
    11·1 answer
  • Experiment 7 static equilibrium of a rigid body.
    14·1 answer
  • Using my raw survival instincts, I am trying to catch a trout from a stream with my bare hands. Despite my exceptional reflexes
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!