Upon what does the energy of a quantum depend?
1 answer:
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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:
a)
b)
Explanation:
Given:
height of water in one arm of the u-tube,
a)
Gauge pressure at the water-mercury interface,:
we've the density of the water
b)
Now the same pressure is balanced by the mercury column in the other arm of the tube:
<u>Now the difference in the column is :</u>
Answer:
V = 1060.8 cm³
Explanation:
we know that the pressure,
P = density x gravity x depth
ρm is the density of mercury
ρw is the density of water
the pressure due to mercury P₁= (ρm) g h₁
the pressure due to water P₂ = (ρw) g h₂
the total pressure
P = P₁ + P₂
P = (ρm) g h₁ + (ρw) g h₂
but the total pressure is double the pressure due to mercury.
(ρm) g h₁ =(ρw) g h₂
h₂ = 81.6 cm
the height of the water is 81.6 cm
the volume
V = height x area
V = 81.6 x 13
V = 1060.8 cm³
the volume of water must be added to double the gauge pressure is V = 1060.8 cm³