Answer:
a) 23.2 e V
b) energy of the original photon is 36.8 eV
Explanation:
given,
energy at ground level = -13.6 e V
energy at first exited state = - 3.4 e V
A photon of energy ionized from ground state and electron of energy K is released.
h ν₁ - 13.6 = K
K combine with photon in first exited state giving out photon of energy
= 26.6 e V
h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°
K + ( 3.4 ) = 26.6 e V
a) energy of free electron
K = 26.6 - 3.4 = 23.2 e V
b) energy of the original photon
h ν₁ - 13.6 = K
h ν₁ = 23.2 + 13.6
= 36.8 e V
energy of the original photon is 36.8 eV
Answer:
h=18.05 cm
Explanation:
Given that
m= 25 kg
K= 1300 N/m
x=26.4 cm
θ= 19.5 ∘
When the block just leave the spring then the speed of block = v m/s
From energy conservation
By putting the values
v=1.9 m/s
When block reach at the maximum height(h) position then the final speed of the block will be zero.
We know that
By putting the values
h=0.1805 m
h=18.05 cm
Gravity affet everything and it touches nothing.
Hope this helps!
Answer: The velocity with which the sand throw is 24.2 m/s.
Explanation:
Explanation:
acceleration due to gravity, a = 3.9 m/s2
height, h = 75 m
final velocity, v = 0
Let the initial velocity at the time of throw is u.
Use third equation of motion
The velocity with which the sand throw is 24.2 m/s.