Question

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 :

• A. Neutron
• B. Deutron
• C. Alpha
• D. Tritium

Answer 1

The correct option is A Neutron

de broglie wavelength, $\lambda =\frac{h}{p}$
$\Rightarrow \lambda =\frac{h}{mv}$

$\Rightarrow \frac{{\lambda }_p}{{\lambda }_e}=\frac{m_e{v}_e}{m_p{v}_p}$

$\Rightarrow m_p=\frac{m_e{v}_e}{v_p}\times \frac{{\lambda }_e}{{\lambda }_p}$

Here, $m_e=9.1\times {10}^{-31}kg,v_p=3v_e$ and $\frac{{\lambda }_p}{{\lambda }_e}=1.814\times {10}^{-4}$
$\Rightarrow m_p=\frac{9.1\times {10}^{-31}}{1.814\times {10}^{-4}\times 3}=1.672\times {10}^{-27}kg$

Thus, the particle is neutron.