Qalaxia Master Bot
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I found an answer from opentextbc.ca

De Broglie's Matter Waves – University Physics Volume 3

Today, this idea is known as de Broglie's hypothesis of matter waves. In 1926, De Broglie's hypothesis, together with Bohr's early quantum theory, ... Any particle that has energy and momentum is a de Broglie wave of frequency f and wavelength \lambda : ... For matter waves, this group velocity is the velocity u of the particle.

For more information, see De Broglie's Matter Waves – University Physics Volume 3

Krishna
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Recall the de Broglie wavelength formula

de Broglie wavelength \lambda = \frac{h}{p}

Where, h - Planck constant = 6.63* 10^{-34} J   and p - linear momentum

Given that

A particle is moving = 3 * electron speed

v = 3 * v_e

\frac{v}{v_e} = 3   .............................(1)

Ratio of de Broglie  Wavelength of the particle to the electron

\frac{\lambda_{\text{ partical }}}{\lambda _{\text{ electron }}} = \frac{\lambda }{\lambda_e} = 1.813*10^{-4}

Mass of an electron m_e = 9.11 × 10^{-31}  kilograms

For a moving particle

mass - m

speed - v

de Broglie wavelength \lambda = \frac{h}{p} = \frac{h}{mv}

Mass m = \frac{h}{\lambda v}

m = \frac{h}{\lambda 3v_e}   ...............................(2)         \because eq(1)

For an electron

mass - m_e

speed - v_e

de Broglie wavelength \lambda_e = \frac{h}{m_e v_e}

Mass m_e = \frac{h}{\lambda_e v_e} .............................(3)

Dividing equation (32) by equation (21)

\frac{eq(3)}{eq(2)}

\frac{m}{m_e}=\frac{\frac{h}{\lambda3v_e}}{\frac{h}{\lambda_ev_e}}

\frac{m}{m_e} = \frac{h}{\lambda (3 v_e)} * \frac{\lambda_e v_e}{h}

\frac{m}{m_e} = \frac{1}{3}* \frac{\lambda_e}{\lambda}

\frac{m}{m_e} = \frac{1}{3} * \frac{1}{1.813*10^{- 4}}

m = m_e * \frac{1}{3} * \frac{1}{1.813*10^{- 4}}

m = (9.11*10^{-31} kg) * \frac{1}{3} * \frac{1}{1.813*10^{- 4}}

m = 1.675 *10^{-27} kg

Hence, the particle may be a proton or a neutron with this mass