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Setler [38]
3 years ago
6

Suppose our apparatus will only allow us to distinguish the first order bright fringe from the central bright spot if they are s

eparated by at least 5º. Find the shortest wavelength of light for which this apparatus is useful
Physics
1 answer:
ozzi3 years ago
3 0

Answer:

λ = 8.716 mm

Explanation:

Given:

- d = 10 cm

- Q >= 5 degrees

Find:

- Find the shortest wavelength of light for which this apparatus is useful

Solution:

- The formula that relates the split difference and angle of separation between successive fringes is given by:

                                            d*sin(Q) = n*λ

Where,

λ: wavelength

d: split separation

Q: angle of separation between successive fringes

m: order number.

- Since this apparatus only shows the first order light so m =1

- the shortest possible wavelength corresponds to:

                                            d*sin(Q) = λ

                                            λ = 0.1*sin(5)

                                            λ = 8.716 mm

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You want to find out how many atoms of the isotope 65Cu are in a small sample of material. You bombard the sample with neutrons
serious [3.7K]

Answer:

a) number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶ atoms

b) m_total Cu = 1.585 10⁹ u = 2.632 10⁻¹⁸ kg

Explanation:

a) For this exercise let's start by using the radioactive decay ratio

           N = N₀  e^{- \lambda t}o e - lambda t

The half-life time is defined as the time it takes for half of the radioactive (activated) atoms to decay, therefore after two half-lives there are

            N = ½ (½ N₀) = ¼ N₀

            N₀ = 4 N

in each decay a photon is emitted so we can use a direct rule of proportions. If an atom emits a photon it has Eo = 1,04 Mev, how many photons it has energy E = 10,000 MeV

          # _atoms = 1 atom (photon) (E / Eo)

          # _atoms = 1 10000 / 1.04

          # _atoms = 9615,4 atoms

          N₀ = 4 #_atoms

          N₀ = 4 9615,4

          N₀=  38461.6  atoms

in the exercise indicates that half of the atoms decay in this way and the other half decays directly to the base state of Zinc, so the total number of activated atoms

          N_activated = 2 # _atoms

          N_activated = 2 38461.6

          N_activated = 76923.2

also indicates that 1% = 0.01 of the nuclei is activated by neutron bombardment

          N_activated = 0.01 N_total

          N_total = N_activated / 0.01

          N_total = 76923.2 / 100

          N_total = 7.692 10⁶ atoms

so the number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶

b) the natural abundance of copper is

  ⁶³Cu     69.17%

  ⁶⁵Cu    30.83%

Let's use a direct proportion rule. If there are 7.692 10⁶  ⁶⁵Cu that represents 30.83, how much ⁶³Cu is there that represents 69.17%

                # _63Cu = 69.17%  (7.692 10⁶    / 30.83%)

                # _63Cu = 17.258 10⁶  atom  ⁶³Cu

the total amount of comatose is

              #_total Cu = #_ 65Cu + # _63Cu

              #_total Cu = (7.692 + 17.258) 10⁶

              #_total Cu = 24.95 10⁶

the atomic mass of copper is m_Cu = 63.546 u

          m_total = #_totalCu m_Cu

          m_total = 24.95 10⁶ 63,546 u

          m_total = 1.585 10⁹ u

let's reduce to kg

           m_total Cu = 1.585 10⁹ u (1,66054 10⁻²⁷ kg / 1 u)

           m_total Cu = 2.632 10⁻¹⁸ kg

8 0
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