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An electron de-excites from the fourth quantum level to the third and then directly to the first. Two frequencies of light are e

4vir4ik [10]

Answer:

The answer is explained below.

Explanation:

The energy emitted during the de-excitation of an electron from a higher energy level to a lower energy level is directly proportional to the frequency of the emitted light.

Here, the total sum of the energies of 2 frequencies of light emitted in different stages is equal to the energy of a single frequency of light during the de-excitation of fourth level to ground level directly.

Hence the total sum of of the frequencies of 2 lights emitted in different stages is equal to the frequency of single frequency of light emitted during the de-excitation from fourth level to ground level directly.

The some of the energies of 2 frequencies emitted by one electron is equal to the energy of a single frequency when electron jumps directly.

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3 years ago

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 15 m/sm/s. A passenger accidentally drops his camera

jeka94

To solve this problem we will apply the linear motion kinematic equations. With the information provided we will calculate the time it takes for the object to fall. From that time, considering that the ascent rate is constant, we will take the reference distance and calculate the distance traveled while the object hit the ground, that is,

$h = v_0 t -\frac{1}{2} gt^2$