The de Broglie wavelength of a particle of mass 1\ {mg} moving with a velocity of 10\ {ms}^{-1} is ( h=6.63 × 10^{-34}\ {J}\ {s}) [AP EAPCET 18 May 2024 Shift 1]

Correct Answer is : 6.63×10−29 m\;6.63 × 10^{-29}\ {m}6.63×10−29 m

Correct Answer is : 6.63×10−29 m\;6.63 × 10^{-29}\ {m}6.63×10−29 m

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Atomic Structure

Section: Chemistry

Question : 32 of 32

Marks: +1, -0

1\ \mathrm{mg}

10\ \mathrm{ms}^{-1}

( h=6.63 \times 10^{-34}\ \mathrm{J}\ \mathrm{s})

Solution:

To calculate the de-Broglie wavelength of a particle, we use the formula:

\lambda = \frac{h}{mv}

where

\lambda

is the wavelength,

h

is Planck's constant,

m

is the mass of the particle, and

v

is the velocity of the particle.

Given:

Mass

m=1\ \mathrm{mg} =1 \times 10^{-3}\ \mathrm{g}=1 \times 10^{-6}\ \mathrm{kg}

Velocity

v=10\ \mathrm{m/s}

Planck's constant

h=6.63 \times 10^{-34}\ \mathrm{Js}

Substitute these values into the formula:

\lambda = \frac{6.63 \times 10^{-34}}{1 \times 10^{-6} \times 10} = \frac{6.63 \times 10^{-34}}{10^{-5}} = 6.63 \times 10^{-29}\ \mathrm{m}

Therefore, the de-Broglie wavelength of the particle is

6.63 \times 10^{-29}\ \mathrm{m}

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