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2 years ago

An electron in a mercury atom drops

1 answer:

4 0

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>

h is Planck constant.

f is photon frequency.

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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If its wavelength were doubled, its energy would be If its wavelength were doubled, its energy would be 4E. 2E. 12E. 14E.

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looking at the formula

energy $E = MC^2$

also, $c = \lambda \times \nu$

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hence, The wavelength was doubled, and its energy will be increased by 4 times.

<h3>What is Wavelength?</h3>

The distance over which a periodic wave's shape repeats is known as the wavelength in physics.
It is a property of both traveling waves and standing waves as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two nearby crests, troughs, or zero crossings.
The spatial frequency is the reciprocal of the wavelength. The Greek letter lambda is frequently used to represent wavelength.
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To learn more about wavelength with the given link

brainly.com/question/13533093

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