Question

Determine the De-Broglie wavelength of a proton, whose kinetic energy is equal to the rest mass energy of an electrons. Given that the mass of the electron is $$9.11 \times 10^{-31} kg$$ and mass of proton is $$1837$$ times as that of the electron.

Answer 1

Here, mass of electron, $$m_0 = 9.1 \times 10^{-31} kg$$
mass of proton, $$m = 1837 \times 9.1 \times 10^{-31} = 1.67 \times 10^{-27} kg$$
Let $$v$$ be the velocity of the proton. Then,
$$\dfrac{1}{2} mv^2 = m_0 c^2$$
or $$m^2 v^2 = 2mm_0c^2$$
or $$mv = \sqrt{2 mm_0 c}$$
$$= \sqrt{2 \times 1.67 \times 10^{-27} \times 9.1 \times 10^{-31}} \times 3 \times 10^8$$
$$= 1.654 \times 10^{-20} kg ms^{-1}$$
Therefore, de Broglie wavelength of the proton,
$$\lambda = \dfrac{h}{mv} = \dfrac{6.62 \times 10^{-34}}{1.654 \times 10^{-20}} = 4 \times 10^{-14} m$$