What is the
(a)    momentum,
(b)    speed, and
(c)    de-Broglie wavelength of an electron with kinetic energy of 120 eV.

Solution

given,
Kinetic energy of the electron, K.E = 120 eV

(a) Momentum of the electron is given as, $p = \sqrt{2 mE}$

$\Rightarrow$ $p = \sqrt{2 \times (9 \times 10^{- 31}) \times (120 \times 1.6 \times 10^{- 19})} = 5.88 \times 10^{- 24} kg {ms}^{- 1}$

(b) Now, since $p = mv$
we have,
$v = \frac{p}{m} = \frac{5.88 \times 10^{- 24}}{9.1 \times 10^{- 31}}$

$\Rightarrow$ $v = 6.46 \times 10^{6} m / s$

(c) De- broglie wavelength of electron is given by,
$λ = \frac{h}{p} = \frac{6.63 \times 10^{- 34}}{5.88 \times 10^{- 24}} = 1.13 \times 10^{- 10} m$

$= 1.13 Å .$

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Wave Optics

What is the
(a)    momentum,
(b)    speed, and
(c)    de-Broglie wavelength of an electron with kinetic energy of 120 eV.

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