Debroglie wavelength of protons accelerated by an electric field at a potential difference V is

Question

Debroglie wavelength of protons accelerated by an electric field at a potential difference V is (A) (0.108/√V) (B) (0.202/√V) (C) (0.286/√V) (D) (0.101/√V)

Tardigrade Admin · Accepted Answer

k ⋅ E=q V=e V=e V P=√2 m k ⋅ E=√2 m e v λ=(h/p)=(h/√2 m e v) (6.62 × 10-34/√2 × 1.67 × 10-27 × 1.681819 × V) λ=(6.62 × 10-34/√3.2 × 1.67 × 10-23) =(6.62/√3.2 × 1.67) × 10-10 × 10-1 =(0.662/√3.2 × 1.67) A° λ=(0.662/2.31 √v) A° =(0.286/√v) A°

Q. Debroglie wavelength of protons accelerated by an electric field at a potential difference V is

$\frac{0.108}{V}$

$\frac{0.202}{V}$

$\frac{0.286}{V}$

$\frac{0.101}{V}$

Q. Debroglie wavelength of protons accelerated by an electric field at a potential difference V is

V​0.108​

V​0.202​

V​0.286​

V​0.101​

k⋅E=qV=eV=eV P=2mk⋅E​=2mev​ λ=ph​=2mev​h​ 2×1.67×10−27×1.681819×V​6.62×10−34​ λ=3.2×1.67​×10−236.62×10−34​ =3.2×1.67​6.62​×10−10×10−1 =3.2×1.67​0.662​A∘ λ=2.31v​0.662​A∘ =v​0.286​A∘

$\frac{0.108}{V}$

$\frac{0.202}{V}$

$\frac{0.286}{V}$

$\frac{0.101}{V}$