When an electron de-excites from the third quantum level to the second, and then to the ground state, two photons are emitted. T
he sum of the emitted frequencies equals the frequency of the single-photon that would be emitted if de-excitation were from the third to ____.
1 answer:
Answer:
ground state
Explanation:
Lets take
n=3 ,n=2 ,n=1 are the energy level.
Energy level n=1 is the ground energy level.
The energy from 3 to 1 = hν
The energy from 3 to 2 = hν₁
The energy from 2 to 1 = hν₂
We can say that
hν = hν₁ + hν₂
If the electron were de-excitation from the third level to ground level then the sum of emitted frequency will be equal to the frequency of a single electron.
Therefore the answer is ground state.
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Answer:
If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2
Explanation:
We know that in a simple harmonic oscillator the maximum speed is given by
=
Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to of the oscillation .
Since
= 2
Where is the maximum speed when frequency is doubled .
Answer:
3.258 m/s
Explanation:
k = Spring constant = 263 N/m (Assumed, as it is not given)
x = Displacement of spring = 0.7 m (Assumed, as it is not given)
= Coefficient of friction = 0.4
Energy stored in spring is given by
As the energy in the system is conserved we have
The speed of the 8 kg block just before collision is 3.258 m/s