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umka21 [38]
2 years ago
12

Why do the lighter isotopes disappear first from the atmosphere? Where do those isotopes go?

Physics
1 answer:
denpristay [2]2 years ago
3 0

Lighter molecules move fast and escape from the upper atmosphere relatively quickly.

To find the answer, we have to know more about the lighter isotopes.

<h3>What are lighter isotopes?</h3>
  • Lighter molecules are mobile and soon leave the higher atmosphere.
  • A particular element's stable isotopes have slightly different atomic masses and quantum mechanical energies.
  • The lighter isotope of an element's chemical bonds are more easily broken than the heavier isotope's.
  • As a result, the light isotope typically benefits from chemical reactions.

Thus, we can conclude that, lighter molecules move fast and escape from the upper atmosphere relatively quickly.

Learn more about the isotopes here:

brainly.com/question/364529

#SPJ4

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the frequency of a beam of uv light is 1.0 ×10 ^15hz what is the energy in one quantum of this light express it in ev? ​
kap26 [50]

Answer:

4.14 eV

Explanation:

f = 1.0 ×10^15 Hz

h= 6.63×10^-34 J s (  this is called PLANCK 'S CONSTANT)

ENEGY = E = ?

E = hf  ( THIS IS FORMULA FOR ENERGY OF ONE QUANTA OR ONE PHOTON )

E= 6.63×10^-34×1.0 ×10^15

E = 6.63×10^-19 J

As 1eV = 1.6×10^-19 J so changing energy in eV from joules we will divide energy by 1.6×10^-19

hence E in eV = 6.63×10^-19/(1.6×10^-19)

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3 years ago
Name the substance in which starch becomes dark blue in colour.<br>​
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3 years ago
A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

   centripetal force acting on satellite = \dfrac{mv^2}{r}

     gravitational force = \dfrac{GMm}{r^2}

    equating both the above equation

    \dfrac{mv^2}{r} = \dfrac{GMm}{r^2}

      v = \sqrt{\dfrac{GM}{r}}

      v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}

          v = 5578.5 m/s

b) T= \dfrac{2\pi\ r}{v}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

          T = 14416.92 s

          T = \dfrac{14416.92}{3600}\ hr

          T = 4 hr

c) gravitational force acting

  F = \dfrac{GMm}{r^2}

  F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}

     F = 5202 N

4 0
3 years ago
Give at least one fact about subatomic particles
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Answer:

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4 0
3 years ago
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