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X-rays of wavelength 'lambda' fall on a ...

X-rays of wavelength `'lambda'` fall on a photosensitive surface emitting electrons. Assuming that the work-function of the surface can be neglected, prove that the de Broglie wavelength of electrons emitted will be `sqrt(( h lambda)/(2mc))`

Text Solution

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`(1)/(2) mv^(2) = hv` or `E_(k) = ( hc)/( lambda)`
or `( p^(2))/( 2m) = ( hc) /( lambda)` or `p = sqrt((2mhc )/( lambda))`
de Broglie wavelength `= ( h )/ (p) = ( h )/( sqrt((2mhc)/( lambda)))`
`= sqrt(( h lambda)/( 2mc))`
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