1.

If kinetic energy of particle is made 16 times,percentage change in its de-Broglie wavelength will be…….

Answer»

25
50
60
75

Solution :de-Broglie wavelength ,
`lambda=(h)/(p)=(h)/(sqrt(2mK))`
`therefore lambda PROP(1)/(sqrt(K))[because h,2,m` are same]
`therefore (lambda_(2))/(lambda_(1))=sqrt((K_(1))/(K_(2)))=sqrt((1)/(16))=(1)/(4)`
`therefore lambda_(2)=(lambda_(1))/(4)`
`therefore` Percentage change,
`=(lambda_(2)-lambda_(1))/(lambda_(1))XX100=((lambda_(1))/(4)-lambda_(1))/(lambda_(1))xx100%`
`=-(3lambda_(1))/(4lambda_(1))xx100%` `therefore` wavelength decrease by 75%


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