Question

An electron with velocity v is found to have a certain value of de Broglie wavelength. The velocity that the neutron should possess to have the same de Broglie wavelength is-

• A. v
• B. v/1840
• C. 1840v
• D. 1840/v

Answer 1

The correct option is B

v/1840

From de-Broglie wavelenght,

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

$\&Weknow$,

$\frac{massofelectron}{massofneutron}$ = $\frac{1}{1840}$

Now,

${\lambda }_e$ = $\frac{h}{m_e{v}_e}$

${\lambda }_n$ = $\frac{h}{m_n{v}_n}$

$\frac{h}{m_e{v}_e}$ = $\frac{h}{m_n{v}_n}$

$v_n$ = $\frac{m_e{v}_e}{m_n}$ = $\frac{1}{1840}$ $\times$ $V_e$ ($m_e$ = $\frac{1}{1840}{m}_n$)