An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is. (nearly): (mt. = 9 × 10-31 kg)

Question

An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is. (nearly): (mt. = 9 × 10-31 kg) (A) 12.2× 1011m (B) 12.2 nm (C) 12.2× 1013m (D) 12.2× 10-12m

Tardigrade Admin · Accepted Answer

For an electron accelerated through a potential V λ = (12.27/√V ) mathring A = (12.27 ×10-10/√10000) = 12.27 ×10-12 m

Q. An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is. (nearly): (mt. = $9 \times 1 0^{- 31}$ kg)

$12.2 \times 1 0^{11}$ m

12.2 nm

$12.2 \times 1 0^{13}$ m

$12.2 \times 1 0^{- 12}$ m

For an electron accelerated through a potential V $λ = \frac{12.27}{V} \overset{˚}{A} = \frac{12.27 \times 1 0 ^{- 10}}{10000} = 12.27 \times 1 0^{- 12} m$

Q. An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is. (nearly): (mt. = 9×10−31 kg)

12.2×1011m

12.2 nm

12.2×1013m

12.2×10−12m

For an electron accelerated through a potential V λ=V​12.27​A˚=10000​12.27×10−10​=12.27×10−12m

Q. An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is. (nearly): (mt. = $9 \times 1 0^{- 31}$ kg)

$12.2 \times 1 0^{11}$ m

$12.2 \times 1 0^{13}$ m

$12.2 \times 1 0^{- 12}$ m

For an electron accelerated through a potential V $λ = \frac{12.27}{V} \overset{˚}{A} = \frac{12.27 \times 1 0 ^{- 10}}{10000} = 12.27 \times 1 0^{- 12} m$