Determine de-Broglie wavelength of an electron having kinetic energy of $1.6 \times 10^{- 6}$ erg.

12.42 A^{o}

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12.42 c m

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12.42 m

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none of these

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Solution

The correct option is C $12.42 c m$
Debroglie wavelength is given by $λ = \frac{h}{m v}$
When KE is given , then $λ = \frac{h}{{(2 m K E)}^{1 / 2}}$

If m= mass of electron in kg , KE in Joules , then answer comes out in Angstrom.
Mass of electron = $9.1 \times 10^{- 31}$ kg
For conversion use :: 1 Joules = $10^{7}$ erg.

Suggest Corrections

Similar questions

1.6 \times 10^{- 6} e r g

(m_{e} = 9.11 \times 10^{- 28} g, h = 6.62 \times 10^{- 27} e r g - s e c)

1 GeV

m = 9.1 \times 10^{- 31} kg, h = 6.6 \times 10^{- 34} Js

e = 1. 6 \times 10^{- 19} C

1.6 \times 10^{- 6} e r g

(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)

1 \times 10^{- 30} k g

= 1.6 \times 10^{- 19} C

200 e V

6.6 \times 10^{- 34} J

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