Answer:
U₂ = 20 J
KE₂ = 40 J
v= 12.64 m/s
Explanation:
Given that
H= 12 m
m = 0.5 kg
h= 4 m
The potential energy at position 1
U₁ = m g H
U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)
U₁ = 60 J
The potential energy at position 2
U₂ = m g h
U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)
U₂ = 20 J
The kinetic energy at position 1
KE= 0
The kinetic energy at position 2
KE= 1/2 m V²
From energy conservation
U₁+KE₁=U₂+KE₂
By putting the values
60 - 20 = KE₂
KE₂ = 40 J
lets take final velocity is v m/s
KE₂= 1/2 m v²
By putting the values
40 = 1/2 x 0.5 x v²
160 = v²
v= 12.64 m/s
Answer:
amount of charge on the source charge
Answer:
λ = 162 10⁻⁷ m
Explanation:
Bohr's model for the hydrogen atom gives energy by the equation
= - k²e² / 2m (1 / n²)
Where k is the Coulomb constant, e and m the charge and mass of the electron respectively and n is an integer
The Planck equation
E = h f
The speed of light is
c = λ f
E = h c /λ
For a transition between two states we have
-
= - k²e² / 2m (1 /
² -1 /
²)
h c / λ = -k² e² / 2m (1 /
² - 1/
²)
1 / λ = (- k² e² / 2m h c) (1 /
² - 1/
²)
The Rydberg constant with a value of 1,097 107 m-1 is the result of the constant in parentheses
Let's calculate the emission of the transition
1 /λ = 1.097 10⁷ (1/10² - 1/8²)
1 / λ = 1.097 10⁷ (0.01 - 0.015625)
1 /λ = 0.006170625 10⁷
λ = 162 10⁻⁷ m
The correct answer should be A