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If velocity of a particle is three times of that of electron and ratio of de Broglie wavelength of particle to that of electron is $1.814 \times 10^{-4}$. The particle will be
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Neutron
de brogile wavelength, $\lambda=\frac{h}{p}$
or $\lambda=\frac{h}{m v}$
$\therefore \quad \frac{\lambda_p}{\lambda_e}=\frac{m_e v_e}{m_p v_p}$
$\Rightarrow \quad m_p=\frac{m_e v_e}{v_p} \times \frac{\lambda_e}{\lambda_p}$
Here, $m_e=9.1 \times 10^{-31} \mathrm{~kg}, v_p=3 v_e$
and $\frac{\lambda_p}{\lambda_e}=1.814 \times 10^{-4}$
$\therefore \quad m_p=\frac{9.1 \times 10^{-31}}{1.814 \times 10^{-4} \times 3}=1.672 \times 10^{-27} \mathrm{~kg}$
Thus, the particle is neutron.
or $\lambda=\frac{h}{m v}$
$\therefore \quad \frac{\lambda_p}{\lambda_e}=\frac{m_e v_e}{m_p v_p}$
$\Rightarrow \quad m_p=\frac{m_e v_e}{v_p} \times \frac{\lambda_e}{\lambda_p}$
Here, $m_e=9.1 \times 10^{-31} \mathrm{~kg}, v_p=3 v_e$
and $\frac{\lambda_p}{\lambda_e}=1.814 \times 10^{-4}$
$\therefore \quad m_p=\frac{9.1 \times 10^{-31}}{1.814 \times 10^{-4} \times 3}=1.672 \times 10^{-27} \mathrm{~kg}$
Thus, the particle is neutron.
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