A metal is irradiated with light of wavelength 600 nm. Given that the work function of the metal is 1.0 eV, the de Broglie wavelength of the ejected electron is close to:

Options

A. `6.6xx10^(-7) m`
B. `8.9xx10^(-11)m`
C. `1.19xx10^(-9) m`
D. `6.6xx10^(-13) m`

Answer

C

Explanation

`E_("absorbed") =(hc)/(lamda) =(6.626xx10^(-34) Js xx 3xx10^(8) ms^(-1))/(600xx10^(-9) m) =0.0331xx10^(-17) J`
`E_("absorbed") =E_(0) +KE, KE = E_("absorbed") -E_(0)=3.31xx10^(-19) J - 1.6 xx10^(-19) J =1.71xx10^(-19) J`
de Broglie wavelength `lamda =(h)/(sqrt(2Em)) = (6.626xx10^(-34) Js)/(sqrt(2xx9.1xx10^(-31)kg xx 1.71xx10^(-19) J))=1.188xx10^(-9) m`