Light of wavelength `500 nm` is incident on a metal with work function `2.28 eV`. The de Broglie wavelength of the emitted electron is

Options

A. `le2.8xx10^(-12)m`
B. `lt2.8xx10^(-10)m`
C. `lt2.8xx10^(-9)m`
D. `ge2.8xx10^(-9)m`

Answer

D

Explanation

`E_(Ph)=(1240)/(500)eV=2.48eV`
`K.E_(max)=E_(Ph)-W`
`=2.48-2.28`
`=0.2eV`
`=0.2eV`
`** lamda_("e min")=(12.27)/(sqrt(K.E_(max)(eV)))`Å
`=(12.27)/(sqrt(0.2))` Å
`=27.436` Å
`=27.436xx10^(-10)m`
`lamda_("min")=2.7436xx10^(-9)m`
`lamda ge lamda_("min")`