1.

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]

Answer»

`le2.8xx10^(-12)m`
`LT 2.8 xx10^(-10)m`
`lt 2.8 xx10^(-9)m`
`ge 2.8 xx10^(-9)m`

Solution :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


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