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o practice Problem-Solving Strategy 39.1: Particles and Waves. A molecule of hydrogen gas has a mass of 3.35×10−27kg3.35×10−27kg and a diameter of 1.48×10−10m1.48×10−10m. What is the kinetic energy at which this molecule's de Broglie wavelength will be equal to its diameter

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ajeigbeibraheem

Answer:

0.0187 eV

Explanation:

Given that:

diameter of the hydrogen gas (λ) = 1.48 ×10⁻¹⁰ m'

mass of the hydrogen gas = 3.35 ×10⁻²⁷ kg

We need to determine the momentum first before calculating the kinetic energy.

So momentum of the hydrogen gas molecule is written as;

[tex]p =\frac{h}{ \lambda}[/tex]

[tex]p = \frac{96.626*10^{-34}J.s}{1.48*10^{-10}m}[/tex]

[tex]p=4.477*10^{-24} kg.m/s[/tex]

NOW, the kinetic energy of the hydrogen gas molecule is calculated as follows by using the formula:

[tex]k_o = \frac{p^2}{2m}[/tex]

[tex]k_o =\frac{(4.477*10^{-24}kg.m/s)^2}{2(3.35*10^-{27})kg}[/tex]

[tex]k_o=2.9916*10^{-21}J(\frac{1eV}{1.6*10^{-19}J})[/tex]

[tex]K_o=0.0187eV[/tex]

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