Question

If the linear momentum of a particle is $2.2\times {10}^{4}kgm{s}^{-1}$, then what will be its de-Broglie wavelength ? (Take hr. $6.6\times {10}^{-34}Js$)

• A. $3\times {10}^{-29}m$
• B. $3\times {10}^{-29}nm$
• C. $6\times {10}^{-29}m$
• D. $6\times {10}^{-29}nm$

Answer 1

Answer: (B)

$3\times {10}^{-29}nm$

Given, the linear momentum of aparticular (p)
$=2.2\times {10}^{4}kgm{s}^{-1}$
$h=6.6\times {10}^{-34}Js$
The de-Broglie wavelength of particle
$\lambda =\frac{h}{p}$
$\lambda =\frac{6.6\times {10}^{-34}}{2.2\times {10}^4}$
or $\lambda =3\times {10}^{-38}m$
or $\lambda =3\times {10}^{-29}nm$