Question

A particle moving with velocity that is three times that of velocity of electron. If ratio of de-Brogile wave length of particle to that of the electron is $1.8\times {10}^{-4}$. Find mass of particle ($m_e=9.1\times {10}^{-31}$ kg )

Answer 1

We use De broglie expression

$\lambda =\frac{h}{mv}$

So we can write,

${\lambda }_e=\frac{h}{m_e{v}_e}$ (1)

${\lambda }_p=\frac{h}{m_p{v}_p}$ (2)

$v_p=3v_e$ (3)

Substituting (3) in (2)

${\lambda }_p=\frac{h}{m_{p}3v_e}$ (4)

Dividing (4) by (3)

$\frac{{\lambda }_p}{{\lambda }_e}=\frac{h}{m_{p}3v_e}.\frac{m_e{v}_e}{h}=1.8\times {10}^{-4}$

$\therefore \frac{m_e}{3m_p}=1.8\times {10}^{-4}$

$\therefore \frac{m_e}{m_p}=5.4\times {10}^{-4}$

$m_p=\frac{9.1\times {10}^{-31}}{5.4\times {10}^{-4}}$

$m_p=1.685\times {10}^{-27}kg$