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ira [324]
3 years ago
6

Two hydrogen atoms collide in a head on collision and end up with zero kinetic energy. Each then emits a photon of 121.6 nm (n=2

to n=1 transition). At what speed were the atoms moving before the collision?
Chemistry
1 answer:
Zinaida [17]3 years ago
8 0

Explanation:

Expression for the kinetic energy is as follows.

         K.E = \frac{1}{2}mv^{2}

Now, total kinetic energy will be as follows.

    K.E = 2 \times \frac{1}{2}mv^{2} = m \times v^{2}

Since, this energy converts into electromagnetic radiation of wavelength 121.6 nm.

Relation between energy and photon is as follows.

   Energy of photon = \frac{hc}{\lambda}

                                = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{121.6 \times 10^{-9}}

                                 = 1.63 \times 10^{-18} J

    m \times v^{2} = 1.63 \times 10^{-18}

          v = \sqrt{\frac{1.63 \times 10^{-18}}{1.67 \times 10^{-27}}

             = 3.12 \times 10^{4} m/s

Thus, we can conclude that atoms were moving at a speed of 3.12 \times 10^{4} m/s before the collision.

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Answer:

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2 years ago
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6 0
3 years ago
7.296 x 10^2 / 9.6 x 10^-9
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Can you use a calculator? If so, demos works great for problems like these!
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