The de Broglie wavelength of an electron with kinetic energy 120 eV is (Given h=6.63×10-34 J s, me=9×10-31 kg, 1 e V =1.6×10-19 J)

Question

The de Broglie wavelength of an electron with kinetic energy 120 eV is (Given h=6.63×10-34 J s, me=9×10-31 kg, 1 e V =1.6×10-19 J) (A) 2.13 mathringA (B) 1.13 mathringA (C) 4.15 mathringA (D) 3.14 mathringA

Tardigrade Admin · Accepted Answer

As p=√2mK =√2×9×10-31×120×1.6×10-19 =5.88×10-24 kg m s-1 de Broglie wavelength, λ=(h/p)=(6.63×10-34/5.88×10-24)=1.13×10-10m =1.13 mathringA

Q. The de Broglie wavelength of an electron with kinetic energy $120 e V$ is (Given $h = 6.63 \times 1 0^{- 34} J s, m_{e} = 9 \times 1 0^{- 31} k g, 1 e V = 1.6 \times 1 0^{- 19} J)$

$2.13 \overset{˚}{A}$

$1.13 \overset{˚}{A}$

$4.15 \overset{˚}{A}$

$3.14 \overset{˚}{A}$

As $p = 2 m K$
$= 2 \times 9 \times 1 0^{- 31} \times 120 \times 1.6 \times 1 0^{- 19}$
$= 5.88 \times 1 0^{- 24} k g m s^{- 1}$
de Broglie wavelength,
$λ = \frac{h}{p} = \frac{6.63 \times 1 0 ^{- 34}}{5.88 \times 1 0 ^{- 24}} = 1.13 \times 1 0^{- 10} m$
$= 1.13 \overset{˚}{A}$

Q. The de Broglie wavelength of an electron with kinetic energy 120eV is (Given h=6.63×10−34Js,me​=9×10−31kg,1eV=1.6×10−19J)

2.13A˚

1.13A˚

4.15A˚

3.14A˚