A metal is irradiated with light of wavelength 660 nm. Given that the work that the work function of the metal is `1.0eV,` the de Broglie wavelength of the ejected electron is close to-

Options

A. `6.6xx10^(-7)m`
B. `8.9xx10^(-11)m`
C. `1.3xx10^(-9)m`
D. `6.6xx10^(-13)m`

Answer

C

Explanation

`E_("Absorbed")=(hc)/(lamda)=(12400)/(600)=1.88eV`
`E_("Absorbed")=E_(0)+KE`
`KE=E_("absorbed")-E_(0)`
`=(1.88-1)eV=0.88eV`
de Broglie wavelength`lamda=(h)/(sqrt(2Em))=sqrt((150)/(KE(eV)))`
`=sqrt((150)/(0.88))=sqrt(170.45)`
`=13"Ã…"=1.3xx10^(-9)m`