Abiotic are nonliving things. So just name 6 nonliving things in finding nemo
a)λ = 1.43 x 10⁻⁷ m, b)λ =5 x 10⁻¹⁹ and, c)λ = 1.99 x 10⁻⁷ m.
<h3>What is an atom?</h3>
- A chemical element is uniquely defined by its atoms, which are tiny pieces of substance. A core nucleus, often surrounded by one or more electrons, makes up an atom. The charge of every electron is negative. The positively charged nucleus has one or more protons and neutrons, relatively heavy particles.
- An atom is referred to any elementary particle of matter with at least one proton. Examples of atoms are neon (N) and hydrogen (H) (Ne).
- Protons, which have a positive charge, and neutrons, which have no charge, make up its structure. All regular, naturally occurring atoms contain the long-lived particles protons, neutrons, and the electrons that orbit them.
- It is difficult to appreciate how little they are until you see how many atoms there are in your body. Around 7 octillion atoms make up one adult.
a) The wavelength of the photon needed to excite an electron from E1 to E4:
ΔE = hc/λ
(-1 + 15) x 10⁻¹⁹ = 6.63 x 10⁻³⁴ x 3 x 10⁸ / λ
λ = 1.43 x 10⁻⁷ m
b) The energy (in joules) a photon must have in order to excite an electron from E2:
= (-5 + 10) x 10⁻¹⁹
λ = 5 x 10⁻¹⁹
c) An electron drops the E3 level to the E1 level, and the atom is said to undergo emission:
Change in energy = 10 x 10⁻¹⁹ J
10 x 10⁻¹⁹ = 6.63 x 10⁻³⁴ x 3 x 10⁸ / λ
λ = 1.99 x 10⁻⁷ m
The wavelength of the photon needed to excite an electron from E1 to E4 is 1.43 x 10⁻⁷ m.
The energy (in joules) a photon must have in order to excite an electron from E2 is 5 x 10⁻¹⁹.
An electron drops the E3 level to the E1 level, and the atom is said to undergo emission is 1.99 x 10⁻⁷ m.
To learn more about the hypothetical atom, refer to:
brainly.com/question/24320731
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A column of elements is a period
1s2 2s2 2p6 3s2 3p1 is the electron configuration of Aluminum, written as Al.
Answer:
40.5% is abundance of X-122
Explanation:
we know there are two naturally occurring isotopes of X, X-120 and X-122.
First of all we will set the fraction for both isotopes
X for the isotopes having mass 122
1-x for isotopes having mass 12
0
The average atomic mass of X is 120.81 amu
we will use the following equation,
122x + 120 (1-x) = 120.81
122x + 120 - 120x = 120.81
122x- 120x = 120.81 -120
2x = 0.81
x= 0.81/2
x= 0.405
0.405 × 100 = 40.5%
40.5% is abundance of X-122 because we solve the fraction x.
now we will calculate the abundance of X-120.
(1-x)
1 - 0.405 = 0.595
0.595 × 100 = 59.5 %
59.5 % for X-120.