Question

if particle of mass 5 mg move with a speed of 8 meter per second then the de broglie wave length will be

Answer 1

12th/Physics

Dual nature of Radiation and Matter

Answer :

Mass of particle = 5mg

(5mg = 5 × 10$^{-6}$kg)

Velocity of particle = 8m/s

We have to find de-beoglie wavelength of particle$.$

_________________________________

◈ De-broglie wavelength associated with object of mass m moving with velocity v is given by

• λ = h / mv

where, h denotes plank's constant

$\dashrightarrow\sf\:\lambda=\dfrac{h}{mv}$

$\dashrightarrow\sf\:\lambda=\dfrac{6.626\times 10^{-34}}{(5\times 10^{-6})(8)}$

$\dashrightarrow\sf\:\lambda=0.165\times 10^{-28}$

$\dashrightarrow\:\boxed{\bf{\purple{\lambda=1.65\times 10^{-29}\:m}}}$

Cheers!

Answer 2

${ \bf{ \underline{ \red{ \underline{ \red{Question}}}} : - }}$

• If particle of mass 5 mg move with a speed of 8 meter per second then the de broglie wave length will be

${ \bf{ \underline{ \red{ \underline{ \red{Solution}}}} : - }}$

{ According to question }

${ \sf{ \dashrightarrow{ \small{ \purple{Mass \: of \: particle \: ( m) }}}}}$

${ \sf{ = \: 5 \: mg \: = 5 \times {10}^{ - 6} \: kg }}$

${ \sf{ \dashrightarrow{ \small{ \purple{velocity \: of \: particle \: (v) }}}} = 8 \:m/s }$

de Broglie wavelength

{de Broglie wavelength :- The wavelength (λ) that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength. A particle’s de Broglie wavelength is usually inversely proportional to its force. }

${ \sf{ \dashrightarrow{ \red{de \: Broglie \: wavelength \: { \bf{\purple{ (\lambda)}}}}}} = \: \frac{h}{mv} }$

where

${ \sf{ \dashrightarrow{\: \: { \purple{Planck’s constant}}}} = \: h}$

${ \sf{ \dashrightarrow{h = \: \: { \purple{6.626× {10}^{ - 34} \: Js}}}} }</p><p>$

{ According to question }

To find de Broglie wavelength

${ \dashrightarrow{ \bf{ \red{ \lambda}}}{ \bf{ \: = \: \frac{h}{mv} }}}$

${ \dashrightarrow{ \bf{ \red{ \lambda}}}{ \bf{ \: = \: \frac{6.626× {10}^{ - 34} }{(5 \times {10}^{ - 6} ) \times( 8)} }}}$

${ \dashrightarrow{ \bf{ \red{ \lambda}}}{ \bf{ \: = \: \frac{6.626× {10}^{ - 34} }{40 \times {10}^{ - 6} } }}}$

${ \dashrightarrow{ \bf{ \red{ \lambda}}}{ \sf{ \: = \: 1.6565\times {10}^{ - 29} \: m}}}</p><p>$

${ \sf{ \underline{Hence, { \red{de \: Broglie \: wavelength }} \: will \: be}}}$

${ \bf{ \red{ \underline{1.6565 \times {10}^{ - 29} \: m.}}}}$

i hope it helps you.