1.

For wave concerned with proton ,de-Broglie wavelength changes by 0.25 %.If its momentum changes by p_(0), initial momentum

Answer»

`100 p_(0)`
`(p_(0))/(400)`
`401 p_(0)`
`(p_(0))/(100)`

Solution :`lambda =(h)/(p)`
EQUATION shows that as p decreases `lambda` increases.
`therefore +(0.25)/(100)lambda=(h)/(p-p_(0))`
`therefore (100.25 lambda)/(100)=(h)/(p-p_(0))` .........(2)
TAKING RATIO of equation (2) and (1),
`(100.25)/(100)=(h)/(p-p_(0))`
`therefore 100.25 p-100.25 p_(0)=100p`
`therefore 0.25 p=100.25 p_(0)`
`therefore p=(100.25 p_(0))/(0.25) therefore p=401 p_(0)`


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