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If travelling at same speeds, which of the following matter waves have the shortest wavelength?
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The correct answer is:
Alpha particle $\left(\mathrm{He}^{2+}\right)$
Alpha particle $\left(\mathrm{He}^{2+}\right)$
From de-Broglie equation, wavelength,
$\lambda=\frac{h}{m v}$
For same speed of different particles, $\lambda \propto \frac{1}{m}$
As $h$ is constant, greater the mass of matter waves, lesser is wavelength and vice-versa. Among these matter waves, alpha particle $\left(\mathrm{He}^{2+}\right)$ has higher mass, therefore, shortest wavelength.
$\lambda=\frac{h}{m v}$
For same speed of different particles, $\lambda \propto \frac{1}{m}$
As $h$ is constant, greater the mass of matter waves, lesser is wavelength and vice-versa. Among these matter waves, alpha particle $\left(\mathrm{He}^{2+}\right)$ has higher mass, therefore, shortest wavelength.
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