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exis [7]
3 years ago
5

A photon ionizes a hydrogen atom from the ground state. The liberated electron 11. recombines with a proton into the first excit

ed state, emitting a 466 A photon. Find a) The energy of the free electron, and b) The energy of the original photon
Physics
1 answer:
anygoal [31]3 years ago
3 0

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

h\nu_2 =\dfrac{hc}{\lambda}=\dfrac{12400}{466}

            = 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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Explanation:

1)

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PE=mgh

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KE=\frac{1}{2}mv^2

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Here Mr. Leppold has both potential and kinetic energy before opening the parachute, because:

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The total mechanical energy of Mr.Leppold is the sum of the potential and the kinetic energy:

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According to the law of conservation of energy, in absence of air resistance, this quantity remains constant.

During the fall, the height of Leppold decreases: this means that as h decreases, the potential energy decreases  too.

However, the total energy E must remain constant: therefore, this means that the kinetic energy KE must increase, and this occurs because the speed of Mr. Leppold increases as he falls.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

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Which of the following has the fewest calence electrons?
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