X-rays of wavelength 0.82Å fall on a metal plate. Find the wavelength associated with photoelectron emitted. Neglect work function of the metal.
Given `c=3xx10^8 m//s h=6.63xx10^(-34)Js.`

Explanation

Here, `lambda=0.82Å =0.82xx10^(-10)m`,
`phi_0=0`
From Einstein's photoelectric equation.
K.E. of electron , `1/2mv^2-=hv-phi_0`
or `mv^2=2hv=(2hc)/lambda ( :' phi_0=0)`
or `mv=sqrt((2hc m)/lambda)`
`:.` de-Broglie wavelength associated with the electron is given by
`lambda'=h/(mv)=h/(sqrt(2hcm//lambda))=sqrt((h lambda)/(2cm))`
`=sqrt((6.6xx10^(-34)xx0.82xx10^(-10))/(2xx3xx10^8xx9.1xx10^(-31))`
`=0.099xx10^(-10)m=0.099Å`