Since the electron dropped from an energy level i to the ground state by emitting a single photon, this photon has an energy of 1.41 × 10⁻¹⁸ Joules.
<h3>How to calculate the photon energy?</h3>
In order to determine the photon energy of an electron, you should apply Planck-Einstein's equation.
Mathematically, the Planck-Einstein equation can be calculated by using this formula:
E = hf
<u>Where:</u>
In this scenario, this photon has an energy of 1.41 × 10⁻¹⁸ Joules because the electron dropped from an energy level i to the ground state by emitting a single photon.
Read more on photons here: brainly.com/question/9655595
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Explanation: The first one
Source: it literally has fusion in the name
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
Answer:
0.657 seconds
Explanation:
speed of wave= wavelength / time period
so
time period= wavelength / speed
= 4.6/7
=0.657 sec
Answer:
motion
Explanation:
the motion causes static which is kinetic energy
