The de Broglie wavelength of an electron with kinetic energy 120 eV is $(G i v e n h = 6.63 \times 10^{- 34} J s, m_{e} = 9 \times 10^{- 31} k g, 1 e V = 1.6 \times 10^{- 19} J)$ :

2.13

˚ A

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1.13

˚ A

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4.15

˚ A

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3.14

˚ A

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Solution

The correct option is D 1.13 $˚ A$
As $p = \sqrt{2 m K . E .}$

$= \sqrt{2 \times 9 \times 10^{- 31} \times 120 \times 1.6 \times 10^{- 19}}$
$= 5.88 \times 10^{- 24} k g m s^{- 1}$

de Broglie wavelength,

$λ = \frac{h}{p} = \frac{6.63 \times 10^{- 34}}{5.88 \times 10^{- 24}} = 1.13 \times 10 - 10 m = 1.13 0 A$

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Similar questions

320 eV

(1 eV = 1.6 \times 10^{- 19} J

= 9.1 \times 10^{- 31} kg

= 6.6 \times 10^{- 34} Js)

80

(1 e V = 1.6 \times 10^{- 19} J,

= 9 \times 10^{- 31} k g

= 6.6 \times 10^{- 34} J - s)

(h = 6.6 \times 10^{- 34} J s, m e = 9.1 \times 10^{- 31} k g)

= 9.1 \times 10^{- 31} k g, h = 6.634 \times 10^{- 34} J s, 1 e V = 1.6 \times 10^{- 19} J

h = 6.63 \times 10^{- 34} J s, m = 9.11 \times 10^{- 31} k g

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