The momentum of a particle having a de-Broglie wavelength of 1017 m is (Given, h = 6.625 × 10-34 m)

Question

The momentum of a particle having a de-Broglie wavelength of 1017 m is (Given, h = 6.625 × 10-34 m) (A) 3.3125 × 10-7 kg m s-1 (B) 26.5 × 10-7 kg m s-1 (C) 6.625 × 10-17 kg m s-1 (D) 13.25 × 10-17 kg m s-1

Tardigrade Admin · Accepted Answer

λ=( h / p ) p =( h /λ) p =(6.625 × 10-34/10-17) p =6.625 × 10-17 kg m s - 1

Q. The momentum of a particle having a de-Broglie wavelength of $1 0^{17} m$ is (Given, $h = 6.625 \times 1 0^{- 34} m$ )

$3.3125 \times 1 0^{- 7} k g m s^{- 1}$

$26.5 \times 1 0^{- 7} k g m s^{- 1}$

$6.625 \times 1 0^{- 17} k g m s^{- 1}$

$13.25 \times 1 0^{- 17} k g m s^{- 1}$

$λ = \frac{h}{p}$
$p = \frac{h}{λ}$
$p = \frac{6.625 \times 1 0 ^{- 34}}{1 0 ^{- 17}}$
$p = 6.625 \times 1 0^{- 17} k g m s^{- 1}$

Q. The momentum of a particle having a de-Broglie wavelength of 1017m is (Given, h=6.625×10−34m)

3.3125×10−7kgms−1

26.5×10−7kgms−1

6.625×10−17kgms−1

13.25×10−17kgms−1