Starlight is released during which nuclear reaction?

Physics

2 answers:

Likurg_2 [28]4 years ago

6 0

Answer:

D. Nuclear fusion

Explanation:

Nuclear fusion is a nuclear reaction where nuclei of atoms( one of low atomic number and another with a higher atomic number) join together to form a much bigger nucleus and release a lot of energy.In the sun, which is a star of course, where light is released,hydrogen is converted to helium and the process releases massive energy into space in form of heat and light.

aliina [53]4 years ago

3 0

Answer:

D. Nuclear fusion

Explanation:

Nuclear fusion is reaction in which two or more atomic nuclei having low atomic numbers join together to form a heavier nucleus usually accompanied with a release of energy.

Stars (like our sun) are very large in size because they are made up of millions of hydrogen and helium atoms. They burn because deep in their core, nuclear fusion reactions occur due to the gases heating up as a result of pressure in the star's core.

This nuclear reaction causes a release of vast amounts of energy in the form of visible electromagnetic radiation. This visible electromagnetic radiation is known as Starlight.

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The star Betelgeuse is 6.1 x 10^18 m away from Earth. How old is the light we see from that star when it reaches us? There are 3

castortr0y [4]

Answer:

635 years old

Explanation:

The light reaching the earth from the sun will travel at a speed called the speed of light, and this has a universal value of 3 × 10⁸ m/s. Bearing this in mind, let us calculate the age of the light reaching the Earth from the sun:

Distance of star from Earth = 6.1 × 10⁸m

Speed of light = 3 × 10⁸ m/s

We have distance and speed, let us calculate the time of travel of the light from the star to the earth.

Distance = speed × time

6.1 × 10⁸ = 3 × 10⁸ × time

$time = \frac{6.1 \times 10^{18}}{3 \times 10^8}$

In order to do the division above, we will divide the whole numbers normally, then we will apply the law of indices to the power that says:

Xᵃ ÷ Xᵇ = X⁽ᵃ⁻ᵇ⁾

$\therefore time = \frac{6.1 \times 10^{18}}{3 \times 10^8}\\= \frac{2.03 \times 10^{(18-8)}}{1} \\= 2.03 \times 10^{10}}\ seconds$

Next, we are told that there are 3.2 × 10⁷ seconds in a year.

∴ The number of years travelled by the light from the star:

$3.2\ \times 10^7\ seconds = 1\ year\\1\ second = \frac{1}{3.2\ \times 10^7} \\\therefore 2.03 \times 10^{10}\ seconds = \frac{2.03 \times 10^{10}}{3.2\ \times 10^7}$