Orbit around hub and arms of spirals ( with name )
1 answer:
You might be interested in
Answer:
λ = 538.0 nm
Explanation:
The solution of the Schrödinger equation for the inner part of the well gives energy
= (h² / 8mL²) n²
Where n is an integer and L is the length of the well
They ask for the transition from the first excited state n = 2 to the base state n = 1
E₂ - E₁ = = (h² / 8mL²) (n₂² - n₁²)
Let's calculate
E₂-E₁ = (6.63 10⁻³⁴)² / (8 9.1 10⁻³¹ (0.7 10⁻⁹)²) (2² -1²)
E₂ –E₁ = 3.6968 10⁻¹⁹ J
Let's use the Planck equation
E = h f
c = λ f
E = h c / λ
E = E₂ ₂- E₁
h c / λ = 3.6968 10⁻¹⁹
λ = h c / (E₂-E₁)
λ = 6.63 10⁻³⁴ 3 10⁸ / 3.6968 10⁻¹⁹
λ = 5.380 10⁻⁷ m
Let's reduce
λ = 5.380 10⁻⁷ m (10 9 nm / 1 m)
λ = 538.0 nm
Answer:
This question is incomplete
Explanation:
The completed question is in the attachment to this question
The two objects (X and Y) at rest had a potential energy of Mx.gH and My.gH respectively because the formula for potential energy (P.E) is mgh. Where m is the mass of the object, g is acceleration due to gravity and h is the height of the object above the ground. Hence, Mx and My are the masses of X and Y respectively.
The change in there potential energy to kinetic energy and back to there potential energy will be
P.E of Y - P.E of X (because mass of Y is larger and will hit the floor first)
Hence, My.gH - Mx.gH ⇔ (My - Mx)gH
Correct option is C
