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murzikaleks [220]
3 years ago
6

Can some please help me with this. Will brainliest

Physics
1 answer:
barxatty [35]3 years ago
8 0

Answer:

1)chest

2)deltoid

3)bicep

4)abs

5)quadriceps

6)lats

7)triceps

8)glutes

9)calves

10)hamstring

11)trapezuis

Explanation:

the explanation is the picture

good luck :)

hopefully, this helps

have a nice day !!

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Ejection of Electrons from Hydrogen by Incident Photons Light of wavelength 80 nm is incident on a sample of hydrogen gas, resul
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Answer:

a)   K_{max} = 1.9 eV = 3.04 10⁻¹⁹ J,b ) This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

Explanation:

a) To calculate the maximum kinetic energy of the expelled electrons let's use the relationships of the photoelectric effect

      K_{max}= h f - Φ

Where K is the kinetic energy, h the Planck constant that is worth 6.63 10⁻³⁴ Js, f the frequency and Φ the work function

The speed of light is related to wavelength and frequency

     c = λ f

Let's analyze the work function, it is the energy needed to start an electron from a metal, in this case to start an electron from a hydrogen atom its fundamental energy is needed, so

     Φ= E₀ = 13.6 eV

let's replace and calculate the energy of the incident photon

     E = h c / λ

     E = 6.63 10⁻³⁴ 3 10⁸/80 10⁻⁹

     E = 2,486 10⁻¹⁸ J

Let's reduce to eV

     E = 2,486 10⁻¹⁸ (1 eV / 1.6 10⁻¹⁹)

     E = 15.5 eV

Now we can calculate the kinetic energy

     K_{max}= h c / f - fi

      K_{max} = 15.5 -13.6

     K_{max} = 1.9 eV

b)     Extra energy = 10.2 eV

The total kinetic energy of electrons is

       Total kinetic energy = 1.9 +10.2 = 12.1 eV

For the calculation we are assuming that all the electors are in the hydrogen base state, but for temperatures greater than 0K some electors may be in some excited state, so less energy is needed to tear them out of hydrogen atom.

Let's analyze this possibility

      ΔE = E photon - Total kinetic energy electron

      ΔE = 15.5 - 12.1

      ΔE = 3.4 eV

If we use the Bohr ratio for the hydrogen atom

     E_{n} = 13.606 / n2

     n = √ 13.606 / En

     n = √ (13606 / 3.4)

     n = 2

This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.

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