A photon ionizes a hydrogen atom from the ground state. The liberated electron 11. recombines with a proton into the first excit
1 answer:
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
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a) -2.54 m/s
b) -2351.25 J
Explanation:
This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum must be equal to the final momentum
:
(1)
Where:
(2)
(3)
is the mass of the first football player
is the velocity of the first football player (to the south)
is the mass of the second football player
is the velocity of the second football player (to the north)
is the final velocity of both football players
With this in mind, let's begin with the answers:
a) Velocity of the players just after the tackle
Substituting (2) and (3) in (1):
(4)
Isolating :
(5)
(6)
(7) The negative sign indicates the direction of the final velocity, to the south
b) Decrease in kinetic energy of the 110kg player
The change in Kinetic energy is defined as:
(8)
Simplifying:
(9)
(10)
Finally:
(10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision