Light of wavelength `5000 Å` falls on a sensitive plate with photoelectric work function of `1.9 eV`. The kinetic energy of the photoelectron emitted will be

Options

A. 0.58 eV
B. 1.24 eV
C. 2.48 eV
D. 1.18 eV

Answer

A

Explanation

Given, wavlength, `lamda=5000` Å`=5xx10^(-7)m`
work function, `W_(0)=1.9eV=1.9xx1.6xx10^(-19)`V
`K_(max)=hv-phi_(0)=(hc)/(lamda)-W_(0)`
`=(6.6xx10^(-34)xx3xx10^(8))/(5xx10^(-7))-1.9xx1.6xx10^(-19)`
`=(6.6xx3)/(5)xx10^(-19)-3.04xx10^(-19)`
`=(3.96-3.04)xx10^(-19)`
`K_(max)~~ 0.92 xx10^(-19)V~~0.58eV`