Light of wavelength 500 nm is incident on a metal with work functin 2.28 eV.The de-Broglie wavelength of the emitted electron is :[`h=6.6xx10^(-34Js)` and `c=3xx10^(8)m//s`]

Options

A. `le2.8xx10^(-12)m`
B. `lt 2.8 xx10^(-10)m`
C. `lt 2.8 xx10^(-9)m`
D. `ge 2.8 xx10^(-9)m`

Answer

D

Explanation

Energy of incident `E=hf=(hc)/(lambda)`
`therefore E=(hc)/(lambda)to` in eV
`=(6.6xx10^(-34)xx3xx10^(8))/(5xx10^(-7)xx1.6xx10^(-19))=2.475`
`~~2.48 eV`
Work function `phi_(0)=2.28eV`
`therefore`According to Einstein.s equation,
`K_(max)=E-phi_(o)=(2.48-2.28)eV` Now, de-Broglie wavelength of electron,
`lambda=(h)/(sqrt(2mE))`
`(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx0.20xx1.6xx10^(-19)))`
`therefore lambda=(6.6xx10^(-34))/(sqrt(5.825xx10^(-50)))`
`therefore lambda=(6.6xx10^(-34))/(2.4132xx10^(-25))`
`therefore =2.788xx10^(-9)` m
`therefore lambda~~28xx10^(-10)m=28Å`
`therefore lambda ge2.8 xx10^(-9)`m