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  • Solve this problem If the kinetic energy of the particle is increased to 16 times its previous value, thepercentage change in the de-Broglie wavelength of the particle is :

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is :

  • Option 1)

    25

  • Option 2)

    75

  • Option 3)

    60

  • Option 4)

    50

 

Answers (2)

best_answer

As we have learned in

De - Broglie wavelength -

\lambda = \frac{h}{p}= \frac{h}{mv}= \frac{h}{\sqrt{2mE}}

- wherein

h= plank's\: constant

m= mass \: of\: particle

v= speed \: of\: the \: particle

E= Kinetic \: energy \: of \: particle

 

 

 

Kinetic energy = \frac{p^{2}}{2m} = \frac{h^{2}}{2m\:\lambda ^{2}}

or \lambda ^{2}= \frac{h^{2}}{2m\:E}         (E = Kinetic energy)

When E becomes 16 E, \lambda becomes \lambda '

\lambda ' \:^{2}= \frac{h^{2}}{2m\left ( 16\:E \right )} = \frac{1}{16}\:\lambda ^{2}

\lambda ' = \frac{\lambda }{4}

Change in wavelength   = \lambda - \lambda ' = \frac{3\lambda }{4}

\therefore \%\: change = \frac{change\:in\:\lambda }{\lambda }\times 100\%=75\%

                          


Option 1)

25

This option is incorrect

Option 2)

75

This option is correct

Option 3)

60

This option is incorrect

Option 4)

50

This option is incorrect

Posted by

Aadil

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2) 75

Posted by

Siddiqui Fareed

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