1.

When de-Broglie wavelength of electron in increased by 1% its momentum…

Answer»

INCREASE by 1%
decrease by 1%
increase by 2%
decreased by 2%

Solution :`lambda=(h)/(p) THEREFORE lambdaprop (1)/(p)therefore lambda prop p^(-1)`
`dlambda prop -p^(2)DP therefore d lambda prop -(1)/(p^(2))dp`
`therefore (dlambda)/(lambda)xx100=-(dp)/(p^(2))xxpxx100`
`therefore (dp)/(p)xx100=-1%`
`therefore` Momentum will decrease by 1%


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