The momentum of a particle which has a de-Broglie wavelength of 0.1 nm is

Question

The momentum of a particle which has a de-Broglie wavelength of 0.1 nm is (A) 3.2 × 10-24 kg ms -1 (B) 4.3 × 10-22 kg ms -1 (C) 5.3 × 10-22 kg ms -1 (D) 6.62 × 10-24 kg ms -1

Tardigrade Admin · Accepted Answer

λ=(h/m v) momentum P=m v ∴ P=(h/λ)=(6.6 × 10-34 kgm 2 s -1/0.1 × 10-9 m ) P=6.6 × 10-24 kg ms -1

Q. The momentum of a particle which has a de-Broglie wavelength of $0.1 nm$ is

$3.2 \times 1 0^{- 24} k g m s^{- 1}$

$4.3 \times 1 0^{- 22} k g m s^{- 1}$

$5.3 \times 1 0^{- 22} k g m s^{- 1}$

$6.62 \times 1 0^{- 24} k g m s^{- 1}$

$λ = \frac{h}{m v}$
momentum $P = m v$
$∴ P = \frac{h}{λ} = \frac{6.6 \times 1 0 ^{- 34} k g m ^{2} s ^{- 1}}{0.1 \times 1 0 ^{- 9} m}$
$P = 6.6 \times 1 0^{- 24} k g m s^{- 1}$

Q. The momentum of a particle which has a de-Broglie wavelength of 0.1nm is

3.2×10−24kgms−1

4.3×10−22kgms−1

5.3×10−22kgms−1

6.62×10−24kgms−1