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. `lt 2.8 xx10^(-9)m`
B. `ge 2.8 xx10^(-9)m`
C. `le 2.8 xx10^(-12)m`
D. `lt 2.8 xx10^(-10)m`

Answer

B

Explanation

`K_("max") = (hc)/lamda -phi_0`
`= (1240)/(500)-2.28 =0.2eV`
`lamda_("min") =(h)/(sqrt(2mK_("max")))=(6.6 xx10^(-34))/(sqrt(2xx9 xx10^(-31) xx0.2 xx1.6 xx10^(-19)))`
`lamda_("min") = 2.8 xx10^(-9)m`