The wavelength of a microscopic particle of mass 9.1 × 10-31 kg is 182 nm, its kinetic energy in J is (h=6.625 × 10-34 J s )

Question

The wavelength of a microscopic particle of mass 9.1 × 10-31 kg is 182 nm, its kinetic energy in J is (h=6.625 × 10-34 J s ) (A) 7.28 × 10-23 (B) 7.28 × 10-24 (C) 3.64 × 10-23 (D) 3.64 × 10-24

Tardigrade Admin · Accepted Answer

Given, mass of particles =9.1 × 10-31 kg Wavelength (λ)=182 nm =182 × 10-9 m According to de-Broglie equation, λ =(h/m v) ⇒ v=(h/m × λ) v =(6.625 × 10-34 Js /9.1 × 10-31 kg × 182 × 10-9 m ) =0.004 × 106 m / s =4 × 103 m / s Therefore, kinetic energy, KE =(1/2) m v2= (1/2) × 9.1 × 10-31 ×(4 × 103)2 =7.28 × 10-24 J

Q. The wavelength of a microscopic particle of mass $9.1 \times 1 0^{- 31} k g$ is $182 nm$ , its kinetic energy in $J$ is $(h = 6.625 \times 1 0^{- 34} J s)$

$7.28 \times 1 0^{- 23}$

$7.28 \times 1 0^{- 24}$

$3.64 \times 1 0^{- 23}$

$3.64 \times 1 0^{- 24}$

Q. The wavelength of a microscopic particle of mass 9.1×10−31kg is 182nm, its kinetic energy in J is (h=6.625×10−34Js)

7.28×10−23

7.28×10−24

3.64×10−23

3.64×10−24

Given, mass of particles =9.1×10−31kg Wavelength (λ)=182nm=182×10−9m According to de-Broglie equation, λ=mvh​ ⇒v=m×λh​ v=9.1×10−31kg×182×10−9m6.625×10−34Js​ =0.004×106m/s=4×103m/s Therefore, kinetic energy, KE=21​mv2=21​×9.1×10−31×(4×103)2 =7.28×10−24J

Q. The wavelength of a microscopic particle of mass $9.1 \times 1 0^{- 31} k g$ is $182 nm$ , its kinetic energy in $J$ is $(h = 6.625 \times 1 0^{- 34} J s)$

$7.28 \times 1 0^{- 23}$

$7.28 \times 1 0^{- 24}$

$3.64 \times 1 0^{- 23}$

$3.64 \times 1 0^{- 24}$