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Allushta [10]
4 years ago
9

A hydrogen atom in a state having a binding energy (the energy required to remove an electron) of -1.51 eV makes a transition to

a state with an excitation energy (the difference between the energy of the state and that of the ground state) of 0 eV.(a) What is the energy of the photon emitted as a result of the transition?
(b) What is the higher quantum number of the transition producing this emission?
(c) What is the lower quantum number of the transition producing this emission?
Physics
1 answer:
jeka944 years ago
6 0

Answer:

a)  ΔE = 1.5 eV , b) n=∞, c) n= 3

Explanation:

a) the energy of the transition is the energy difference between the two states

         ΔE = E₂ –E₁

         ΔE = 0 - (-1.5)

         ΔE = 1.5 eV

b) The energy of each state is given by the equation

           E = -13.606 / n²

The highest quantum number is

         n = √ (13,606 / E)

        n = √ (13.606 /1.5)

        n = 3

the transition is  n = ∞  that has energy 0

c) the lowest quantum number is 3

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A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.
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Given Information:

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b) Resistance = R = 484 Ω

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According to the Ohm’s law, the power dissipated in the light bulb is given by

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3 years ago
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3 years ago
A flat coil of wire has an inductance of 40. 0 mh and a resistance of 6. 00 ω. It is connected to a 21. 2-v battery at the insta
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For a  flat coil of wire has an inductance of 40. 0 mh and a resistance of 6. 00 ω, the rate of energy being delivered is mathematically given as

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