An electron is moving with a kinetic energy of 4.55 × 10- 25 J. The de Broglie wavelength for it is (Given: texth = 6.626 × 10- 34 , textm texte = 9.1 × 10- 31 textkg )

Question

An electron is moving with a kinetic energy of 4.55 × 10- 25 J. The de Broglie wavelength for it is (Given: texth = 6.626 × 10- 34 , textm texte = 9.1 × 10- 31 textkg ) (A) 7.2× 10- 7m (B) 72× 10- 7m (C) 0.72× 10- 7m (D) 4.2× 10- 7m

Tardigrade Admin · Accepted Answer

textK text.E text. = (1/2) textmv text2 V=√(2 K E/m) =√(2 × 4.55 × 1 0- 25/9.1 × 1 0- 31 k g) = 103 textmS- 1 λ =(h/m v)=(6.626 × 1 0- 34/9.1 × 1 0- 31 × 1 03) = 7.2 × 10- 7 textm

Q. An electron is moving with a kinetic energy of 4.55 $\times 1 0^{- 25}$ J. The de Broglie wavelength for it is
(Given: $h = 6.626 \times 1 0^{- 34}, m_{e} = 9.1 \times 1 0^{- 31} kg$ )

$7.2 \times 1 0^{- 7} m$

$72 \times 1 0^{- 7} m$

$0.72 \times 1 0^{- 7} m$

$4.2 \times 1 0^{- 7} m$

$K .E . = \frac{1}{2} mv^{2}$
$V = \frac{2 K E}{m}$
$= \frac{2 \times 4.55 \times 1 0 ^{- 25}}{9.1 \times 1 0 ^{- 31} k g}$
$= 1 0^{3} mS^{- 1}$
$λ = \frac{h}{m v} = \frac{6.626 \times 1 0 ^{- 34}}{9.1 \times 1 0 ^{- 31} \times 1 0 ^{3}}$
$= 7.2 \times 1 0^{- 7} m$

Q. An electron is moving with a kinetic energy of 4.55 ×10−25 J. The de Broglie wavelength for it is (Given: h=6.626×10−34,me​=9.1×10−31kg )

7.2×10−7m

72×10−7m

0.72×10−7m

4.2×10−7m