The de-Broglie wavelength of an electron of kinetic energy 9 eV is (take, h=4 × 10-15 eV - s , c=3 × 1010 cm / s and the mass me of electron as .me c2=0.5 MeV )

Question

The de-Broglie wavelength of an electron of kinetic energy 9 eV is (take, h=4 × 10-15 eV - s , c=3 × 1010 cm / s and the mass me of electron as .me c2=0.5 MeV ) (A) 4 × 10-8 cm (B) 3 × 10-8 cm (C) 4 × 10-7 cm (D) 3 × 10-7 cm

Tardigrade Admin · Accepted Answer

For electron, p c=√2 KE m0 c2 ⇒ p c= √2 × 9 eV × 0.58 × 106 ⇒ p c=3 × 103 eV Now, de-Broglie wavelength, λ=(h c/p c)=(4 × 10-15 × 3 × 1010/3 × 103)=4 × 10-8 cm (∵ c=3 × 1010 cm / s )

Q. The de-Broglie wavelength of an electron of kinetic energy $9 e V$ is (take, $h = 4 \times 1 0^{- 15} e V - s, c = 3 \times 1 0^{10} c m / s$ and the
mass $m_{e}$ of electron as $m_{e} c^{2} = 0.5 M e V)$

$4 \times 1 0^{- 8} c m$

$3 \times 1 0^{- 8} c m$

$4 \times 1 0^{- 7} c m$

$3 \times 1 0^{- 7} c m$

Q. The de-Broglie wavelength of an electron of kinetic energy 9eV is (take, h=4×10−15eV−s,c=3×1010cm/s and the mass me​ of electron as me​c2=0.5MeV)

4×10−8cm

3×10−8cm

4×10−7cm

3×10−7cm

For electron, pc=2KEm0​c2​ ⇒pc=2×9eV×0.58×106​ ⇒pc=3×103eV Now, de-Broglie wavelength, λ=pchc​=3×1034×10−15×3×1010​=4×10−8cm (∵c=3×1010cm/s)

Q. The de-Broglie wavelength of an electron of kinetic energy $9 e V$ is (take, $h = 4 \times 1 0^{- 15} e V - s, c = 3 \times 1 0^{10} c m / s$ and the
mass $m_{e}$ of electron as $m_{e} c^{2} = 0.5 M e V)$

$4 \times 1 0^{- 8} c m$

$3 \times 1 0^{- 8} c m$

$4 \times 1 0^{- 7} c m$

$3 \times 1 0^{- 7} c m$