The kinetic energy of an electron is 5 eV. Calculate the de-Broglie wavelength associated with it ( h=6.6 × 10-34 Js, .me=9.1 × 10-31 kg )

Question

The kinetic energy of an electron is 5 eV. Calculate the de-Broglie wavelength associated with it ( h=6.6 × 10-34 Js, .me=9.1 × 10-31 kg ) (A) 5.47 mathringA (B) 10.9 mathringA (C) 2.7 mathringA (D) None of these

Tardigrade Admin · Accepted Answer

Wavelength associated with an electron λ =(h/√2 m E) =(6.6 × 10-34/√2 × 9.1 × 10-31 × 5 × 1.6 × 1019) =5.47 × 10-10 m =5.47 mathringA

Q. The kinetic energy of an electron is $5 e V$ . Calculate the de-Broglie wavelength associated with it ( $h = 6.6 \times 1 0^{- 34} J s$ , $m_{e} = 9.1 \times 1 0^{- 31} k g)$

$5.47 \overset{˚}{A}$

$10.9 \overset{˚}{A}$

$2.7 \overset{˚}{A}$

None of these

Wavelength associated with an electron
$λ = \frac{h}{2 m E}$
$= \frac{6.6 \times 1 0 ^{- 34}}{2 \times 9.1 \times 1 0 ^{- 31} \times 5 \times 1.6 \times 1 0 ^{19}}$
$= 5.47 \times 1 0^{- 10} m = 5.47 \overset{˚}{A}$

Q. The kinetic energy of an electron is 5eV. Calculate the de-Broglie wavelength associated with it ( h=6.6×10−34Js, me​=9.1×10−31kg)

5.47A˚

10.9A˚

2.7A˚