Which is not a reason why the mass of an atom is expressed in atomic mass units rather than in grams?
2 answers:
Answer:
Ease of use and readability.
Explanation:
An atom of Oxygen, as an example, weighs 0.000...0001661 grams (with 23 zeros in front of "1661". In scientific notation this is written more concisely as
Still, in both cases the representation in units of grams is not easily readable and inconvenient to handle, especially when used in equations multiplying such mass with other measures. For this reason, the mass was agreed to be measured in atomic units, or a.u., which offer a numerically comparable baseline. An atomic unit for mass relates to the mass of an electron which is small enough to provide a good measurement unit. There are several variations to this so I recommend further reading if you need more detail.
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Answer:
rom level n = 7 to level n = 3
Explanation:
Bohr's model describes the energy levels for the hydrogen atom
En = -13.606 / n²
Where n is an integer with values of 1, 2, 3
An electronic transition occurs between two permitted levels of energy
ΔE = -
Let's apply these relationships our problem.
Let's start by knowing the energy of level n = 7
E₇ = - 13.606 / 7²
E₇ = - 0.27767 eV
Now let's see what the energy of the emitted photon
E = h f
c = λ f
f = c / λ
E = h c / λ
E = 6.63 10⁻³⁴ 3 10⁸/1005 10⁻⁹
E = 19,791 10⁻²⁰ J
Let's reduce to eV
E = 19,791 10⁻²⁰ (1 eV / 1.6 10⁻¹⁹)
E = 1,237 eV
The possible transitions from this level are towards n = 6, 5, 4,3,2, 1
We must test the different values until we find the right one
Energy of the states
n
6 -0.378
5 -0.544
4 -0.850
3 -1,512
2 -3,402
1 -13,606
Let's examine the transition n = 7 to n = 6
ΔE = - 0.27767 - (-0.3779)
ΔE = 0.10023 eV
n = 7 to n = 5
ΔE = -0.27767 - (-0.5442)
ΔE = 0.267 eV
n = 7 a n = 4
ΔE = -0.27767- (-0.8504)
ΔE = 0.573 eV
n = 7 a n = 3
ΔE = -0.27767 - (- 1.5118)
ΔE = 1.234 Ev
This is the transition sought, so that the electron goes from level n = 7 to level n = 3