An electron has kinetic energy 2.8 × 10-23 J. de-Broglie wavelength will be nearly (me= 9.1 × 10-31 kg)

Question

An electron has kinetic energy 2.8 × 10-23 J. de-Broglie wavelength will be nearly (me= 9.1 × 10-31 kg) (A) 9.28 × 10-4 m (B) 9.28× 10-7 m (C) 9.28 ×10-8 m (D) 9.28 × 10-10 m

Tardigrade Admin · Accepted Answer

Formula for de-Broglie wavelength is λ=(h/p) or λ =(h/mv) ⇒ eV =(1/2) mv2 or v=√(2eV/m) λ=(h/√2me V)=(6.62×10-34/√2×9.1×10-31×2.8×10-23) λ =9.28 ×10-8 meter.

Q. An electron has kinetic energy $2.8 \times 1 0^{- 23}$ J. de-Broglie wavelength will be nearly $(m_{e} = 9.1 \times 1 0^{- 31}$ kg)

$9.28 \times 1 0^{- 4} m$

$9.28 \times 1 0^{- 7} m$

$9.28 \times 1 0^{- 8} m$

$9.28 \times 1 0^{- 10} m$

Formula for de-Broglie wavelength is
$λ = \frac{h}{p} or λ = \frac{h}{m v} \Rightarrow e V = \frac{1}{2} m v^{2} or v = \frac{2 e V}{m}$
$λ = \frac{h}{2 m e V} = \frac{6.62 \times 1 0 ^{- 34}}{2 \times 9.1 \times 1 0 ^{- 31} \times 2.8 \times 1 0 ^{- 23}}$
$λ = 9.28 \times 1 0^{- 8} m e t er .$

Q. An electron has kinetic energy 2.8×10−23 J. de-Broglie wavelength will be nearly (me​=9.1×10−31 kg)

9.28×10−4m

9.28×10−7m

9.28×10−8m

9.28×10−10m