Answer:
The initial energy level = 6
Explanation:
Photon wavelength is proportional to energy. The wavelength of emitted photons is related to the energy levels of the atom as given by the Rydberg formula:
ₕ₁₂
(1/λ) = Rₕ [(1/n₂²) − (1/n₁²)]
where n₂ = final energy level = 2
n₁ = initial energy level = ?
Rₕ = Rydberg's constant = 1.097 × 10⁷ m⁻¹
λ = wavelength = 410 nm = 410 × 10⁻⁹ m
1/(410 × 10⁻⁹) = (1.097 × 10⁷) [(1/2²) − (1/n₁²)]
0.223 = [(1/4) − (1/n₁²)]
(1/n₁²) = 0.02778
n₁² = 1/0.02778 = 36
n₁ = 6.
Answer: Dark matter.
Explanation: Hope it helps :)
Answer:
discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC source
Explanation:
Bulbs can emit light in several ways:
* When the emission is carried out by the heating of its filament, the bulb is called incandescent, in general its spectrum is similar to that of a black body, this is a continuous spectrum with a maximum dependent on the fourth power of the temperature of the filament.
* The emission can be by atomic transitions, in this case there is a discrete spectrum formed by the spectral lines of the material that forms the gas of the lamp, in general for the yellow emission the most used materials are mercury and sodium or a mixture of they.
Consequently, as discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC type
Answer:
60mph=26.8224meters per second
Explanation:
Answer:
v’= 9.74 m / s
Explanation:
The Doppler effect is due to the relative movement of the sound source and the receiver of the sound, in this case we must perform the exercise in two steps, the first to find the frequency that the bat hears and then the frequency that the audience hears that also It is sitting.
Frequency shift heard by the murciela, in case the source is still and the observer (bat) moves closer
f₁ ’= f₀ (v + v₀)/v
Frequency shift emitted by the speaker in the bat, in this case the source is moving away from the observer (public sitting) that is at rest
f₂’= f₁’ v/(v - vs)
Note that in this case the bat is observant in one case and emitter in the other, called its velocity v’
v’= vo = vs
Let's replace
f₂’= f₀ (v + v’)/v v/(v -v ’)
f₂’= f₀ (v + v’) / (v -v ’)
(v –v’ ) f₂’ / f₀ = v + v ’
v’ (1+ f₂’ /f₀) = v (f₂’/fo - 1)
v’ (1 + 1.059) = 340 (1.059 - 1)
v’= 20.06 / 2.059
v’= 9.74 m / s
