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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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Explanation:

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Which profile best describes the reaction C(s) + 2H2(g) → CH4(g),<br> AH = -74.9 kJ?
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Answer:

Option B. A

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From the reaction above, we can see that the enthalpy change (ΔH) is negative (i.e –74.9 KJ) which implies that the heat content of the reactants is greater than the heat content of the products. Thus, the reaction is exothermic reaction.

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