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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.

Answer

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.
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