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A particle of mass 1mg has the same wavelength as the electron moving with a velocity of $3 \times {10^6}m/s$ . The velocity of the particle is (mass of electron= $9.1 \times {10^{ - 31}}kg$ )
A. $2.7 \times {10^{ - 23}}m/s$
B. $2.7 \times {10^{ - 18}}m/s$
C. $9 \times {10^{ - 2}}m/s$
D. $3 \times {10^{ - 31}}m/s$

Last updated date: 11th Jun 2024
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Hint: We have applied the concept of de Broglie hypothesis which states that wavelength of a particle is inversely proportional to its momentum.Therefore, the smaller mass and smaller momentum of the electron means that it has a longer wavelength.
Mathematically, $\lambda \propto \dfrac{1}{p} \Rightarrow \lambda = h.\dfrac{1}{p}$ where $h$ is the Planck’s constant.

Complete step by step answer:
We write the wavelength of a particle as,
$\lambda = \dfrac{h}{p} = \dfrac{h}{{mv}}$
where $\lambda $ is the wavelength of a particle, $h$ is the Planck’s constant, $p$ is the momentum of the particle, $m$ is the mass of the particle and $v$ is the velocity of the particle. Let the wavelength of the given particle be $\lambda $ and the wavelength of an electron be ${\lambda _e}$

We assume the momentums, masses and velocities of the particle and electron as $\left( {p,{p_e}} \right),\left( {m,{m_e}} \right)\& \left( {v,{v_e}} \right)$ respectively
According to the question $\lambda = {\lambda _e}$
\dfrac{h}{p} = \dfrac{h}{{{p_e}}} \\
\Rightarrow p = {p_e} \\
\Rightarrow mv = {m_e}{v_e} \\
\Rightarrow v = \dfrac{{{m_e}{v_e}}}{m} \\
Substituting the given data,
{m_e} = 9.1 \times {10^{31}}kg,{v_e} = 3 \times {10^6}m/s, \\
\Rightarrow m = 1mg = 1 \times {10^{ - 6}}kg \\
\Rightarrow v = \dfrac{{9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^6}}}{{1 \times {{10}^{ - 6}}}} \\
\therefore v = 2.7 \times {10^{ - 18}}m/s \\
Therefore, the correct answer is option B.

Note:We can also solve for the wavelength of the electron first using the value of Planck’s constant and then equate it to the wavelength of the given particle. Also, we can get a rough idea without solving for the exact answer if the options are at a wide range because from the formula we can see that The longer de Broglie wavelength of particle 1 means that it has less momentum than particle 2 because de Broglie wavelength is inversely proportional to momentum. And if the two particles have the same mass, this also means particle 1 has a smaller velocity and less kinetic energy than particle 2.