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zheka24 [161]
2 years ago
5

A 70.-kg person exposed to ⁹⁰Sr absorbs 6.0X10⁵ β⁻ particles, each with an energy of 8.74X10⁻¹⁴ J.(b) If the RBE is 1.0, how man

y millirems is this?
Chemistry
1 answer:
Tpy6a [65]2 years ago
6 0

A 70.-kg person exposed to ⁹⁰Sr absorbs 6.0X10⁵ β⁻ particles, each with an energy of 8.74X10⁻¹⁴ J.

<h3>What is β⁻ particles ?</h3>

A beta particle, also known as a beta ray or beta radiation (symbol ), is a highly energetic, swiftly moving electron or positron that is released during the radioactive disintegration of an atomic nucleus. Beta decay occurs in two ways: decay and + decay, which result in the production of electrons and positrons, respectively.

In air, beta particles with an energy of 0.5 MeV have a range of roughly one meter; the range is energy-dependent.

Ionizing radiation of the sort known as beta particles is regarded, for the purposes of radiation protection, as being more ionizing than gamma rays but less ionizing than alpha particles. The damage to live tissue increases as the ionizing effect increases, but so does the radiation's penetration power.

To learn more about β⁻ particles from the given link:

brainly.com/question/10111545

#SPJ4

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What is the mass percentage of silicon?​
Mkey [24]

Answer: 46.743%

Explanation: what is the mass percentage of silicon?​ 46.743%

***If you found my answer helpful, please give me the brainliest, please give a nice rating, and the thanks ( heart icon :) ***

8 0
3 years ago
Give the n and l values and the number of orbitals for sublevel 5g.
Pepsi [2]

The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

There are total four quantum numbers:

1) Principal quantum number , n

2) Angular quantum number , l

3) Magnetic quantum number , ml

4) spin quantum number , ms

For 5g shell, n = 5

subshell g , l = 4     ....0 - s , 1 - p , 2 - d, 3 - f, 4 -g

number of orbitals in subshell = (2l + 1)  ( 2×4 + 1) = 9

Thus,  The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

To learn more about quantum numbers here

brainly.com/question/14650894

#SPJ1

6 0
1 year ago
it takes 463 kJ/mol to break an oxygen-hydrogen single bond. calculate the maximum wavelength of light for which an oxygen-hydro
Marianna [84]

Answer:

  • <u>2.59 × 10⁻⁷ m  = 259 nm</u>

Explanation:

You need to calculate the wavelength of a photon with an energy equal to 463 kJ/mol, which is the energy to break an oxygen-hydrogen atom.

The energy of a photon and its wavelength are related by the Planck - Einstein equation:

  • E = hν,

Where:

  • E = energy of a photon,
  • h = Planck constant (6.626 × 10⁻³⁴ J . s) and
  • ν = frequency of the photon.

And:

  • ν = c / λ

Where:

  • c = speed of light (3.00 × 10⁸ m/s in vacuum)
  • λ = wavelength of the photon

Thus, you can derive:

  • E = h c / λ

Solve for λ:

  • λ = h c / E

Before substituting the values, convert the energy, 463 kJ/ mol, to J/bond

  • 463 kJ/ mol × 1,000 J/kJ × 1 mol / 6.022 × 10 ²³ atom × 1 bond / atom

       = 7.69×10²³ J / bond

Substitute the values and use the energy of one bond:

  • λ = 6.626 × 10⁻³⁴ J . s × 3.00 × 10⁸ m/s / 7.69×10²³ J = 2.59 × 10⁻⁷ m

The wavelength of light is usually shown in nanometers:

  • 2.59 × 10⁻⁷ m × 10⁹ nm / m = 259 nm ← answer
6 0
3 years ago
What is the kinetic energy of a 1-kilogram ball is thrown into the air with ab ititial velocity of 30m/s?
mariarad [96]
KE=1/2*mass*velocity^2
So u do 1/2 * 1 * 30^2
1/2 * 1 * 900
= 450kgm/s

P.s. I'm not sure if I would have to convert kg to g.
Anyways hope this helped
4 0
3 years ago
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
4 years ago
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