Question

The momentum of a particle having a de-Broglie wavelength of $10^{17}\, m$ is (Given, $h = 6.625 \times 10^{-34} \,m$)

• A. $3.3125 \times 10^{-7} \,kg\, m\, s^{-1}$
• B. $26.5 \times 10^{-7} \,kg \,m\, s^{-1}$
• C. $6.625 \times 10^{-17}\, kg \,m\, s^{-1}$
• D. $13.25 \times 10^{-17} \,kg \,m\, s^{-1}$

Answer 1

Answer: (C)

$\lambda=\frac{ h }{ p }$
$p =\frac{ h }{\lambda}$
$p =\frac{6.625 \times 10^{-34}}{10^{-17}}$
$p =6.625 \times 10^{-17} \,kg \,m s ^ { - 1 }$