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3 years ago

What happens when an electron moves from an excited state to the ground state?

Physics

1 answer:

Darina [25.2K]3 years ago

6 0

<span>When an electron moves from an excited state to the ground state, "Energy releases"

Hope this helps!</span>

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A strontium vapor laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an a

sladkih [1.3K]

Answer:

$d=1.29*10^{-6}m$

Explanation:

From the question we are told that:

Distance of wall from CD $D=1.4$

Second bright fringe $y_2= 0.803 m$

Let

Strontium vapor laser has a wavelength \lambda= 431 nm=>431 *10^{-9}m

Generally the equation for Interference is mathematically given by

$y=frac{n*\lambda*D}{d}$

Where

$d=\frac{n*\lambda*D}{y}$

$d=\frac{2*431 *10^{-9}m*1.4}{0.803}$

$d=1.29*10^{-6}m$

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3 years ago

Which term defines the energy of motion?

Korvikt [17]

D kinetic energy✨
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3 years ago

Pls answer quick will mark brainlest and 5 stars

QveST [7]

Answer:A

Explanation:

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3 years ago

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Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km fro

luda_lava [24]

Answer:

c. $4.582\times10^{21} kg$

Explanation:

$r_{i}$ = Initial distance between asteroid and rock = 7514 km = 7514000 m

$r_{f}$ = Final distance between asteroid and rock = 2823 km = 2823000 m

$v_{i}$ = Initial speed of rock = 136 ms⁻¹

$v_{f}$ = Final speed of rock = 392 ms⁻¹

$m$ = mass of the rock

$M$ = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

$(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg$

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3 years ago

To solve a problem using the equation for keplerâs third law, enrico must convert the average distance of mars from the sun from

Natali5045456 [20]

1 astronomical unit = 149597870700m
Enrico should divide distance in meters with this number.

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3 years ago

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