Question

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Answer 1

0.364 × 10^-9 m is the required answer.

GIVEN:

• Velocity of electron , v = 2 × 10⁶ m/s

• Mass of electron, m = 9.1 × 10^-31 kg

• Planck's constant , h = 6.625 × 10^-34 Js

TO FIND:

De broglie wavelength , λ

FORMULA:

According to de broglie's equation,

de broglie wavelength , λ= h/mv

SOLUTION:

$wavelength, \:\lambda = \frac{h}{mv} \\ \\ = \frac{6.625 \times {10}^{ - 34} }{9.1 \times {10}^{ - 31} \times 2 \times {10}^{6} } \\ \\ = \frac{6.625 \times {10}^{ - 34} \times {10}^{31} \times {10}^{ - 6} }{9.1 \times 2} \\ \\ = \frac{6.625 \times {10}^{ - 34 + 31 - 6} }{18.2} \\ \\ = \frac{6.625 \times {10}^{ - 9} }{18.2} \\ \\ \lambda\:= 0.364 \times {10}^{ - 9} m$

ANSWER:

De broglie wavelength of electron is 0.364 × 10^-9 m

DE BROGLIE EQUATION:

• It proves the wave nature of particle.

• Since the wavelength is inversely proportional to mass of the object, de broglie wavelength is appreciable in case of objects of very small mass.

• Since the wavelength is inversely proportional to velocity, the velocity decreases with increase in wavelength.

Answer 2

Answer:

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