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## QuestionPhysicsClass 12

A particle is moving three times as fast as an electron.The ratio of the de Broglie wavelength of the particle to that of theelectron is $$1.813\times 10^{-4}$$ . Calculate the particle’s mass and identify the particle.

According to the de Broglie equation,
$$\lambda =\frac h p$$
Planck’s constant, $$h=6.62\times 10^{-34} Js$$
For a particle having mass $$m$$  and velocity $$v$$ .
$$\lambda =\frac h{\mathit{mv}}$$
$$m=\frac h{\mathit{\lambda v}}$$
For an electron,
$$m_e=\frac h{\lambda _ev_e}$$
$$\frac{\lambda }{\lambda _e}=1.813\times 10^{-4}$$
$$\frac v{v_e}=3$$
Mass of the particle,
$$m=m_e\times \frac{\lambda }{\lambda _e}\times \frac v{v_e}$$
$$m=9.11\times 10^{-31}\times \frac 1{1.813\times 10^{-4}}\times \frac 1 3=1.675\times 10^{-27}\mathit{kg}$$
This mass could be of a proton or a neutron.