Answer:
Explanation:
Atoms—and the protons, neutrons, and electrons that compose them—are extremely small. For example, a carbon atom weighs less than 2 × 10−23 g, and an electron ... The amu was originally defined based on hydrogen, the lightest element, ... but three-letter symbols have been used to describe some elements that have ...
Protons: Protons are positively charged particles that are also found in the nucleus. Like neutrons, protons give mass to the atom but do not participate in ... 3) Electrons: Electrons are negatively charged particles that are found in ... pair of electrons with 4 different hydrogen atoms, forming a molecule of CH4 (methane).Elements differ from each other in the number of protons they have, e.g. ... Atoms of an element that have differing numbers of neutrons (but a constant atomic ... Electrons, because they move so fast (approximately at the speed of light), ...toms are made up of particles called protons, neutrons, and electrons, which ... Therefore, they do not contribute much to an element's overall atomic mass. ... For instance, iron, Fe, can exist in its neutral state, or in the +2 and +3 ionic states. ... Isotopes of the same element will have the same atomic number but different ...
Answer:
a) m =1 θ = sin⁻¹ λ / d, m = 2 θ = sin⁻¹ ( λ / 2d)
, c) m = 3
Explanation:
a) In the interference phenomenon the maxima are given by the expression
d sin θ = m λ
the maximum for m = 1 is at the angle
θ = sin⁻¹ λ / d
the second maximum m = 2
θ = sin⁻¹ ( λ / 2d)
the third maximum m = 3
θ = sin⁻¹ ( λ / 3d)
the fourth maximum m = 4
θ = sin⁻¹ ( λ / 4d)
b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.
I = I₀ cos² (Ф) (sin x / x)²
Ф = π d sin θ /λ
x = pi a sin θ /λ
where a is the width of the slits
with the values of part a are introduced in the expression and we can calculate intensity of each maximum
c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present
maximum interference d sin θ = m λ
first diffraction minimum a sin θ = λ
we divide the two expressions
d / a = m
In our case
3a / a = m
m = 3
order three is no longer visible
