If a light of wavelength 330nm is incident on a metal with word function 3.55eV, the electrons are emitted. Then the wavelength of the emitted electron is ( Taken `h = 6.6 xx 10^(-34) Js)`

Options

A. `lt 2.75 xx 10^(-9) m `
B. `ge 2.75 xx 10^(-9) m`
C. `le 2.75 xx 10^(-12) m`
D. `lt 2.5 xx 10^(-10) m`

Answer

A

Explanation

Maximum KE of emitted electron is
`K_("max")=(hc)/(lambda)-phi_0 = ((1240)/(330)-3.55) eV= (3.76 - 3.55) eV`
`K_("max")=0.21 eV`
de-Broglie wavelength of emitted electron
`lambda=(h)/(sqrt(2mKE))=(6.63 xx 10^(-34))/(sqrt(2 xx 9.1 xx 10^(31) xx 0.21 xx 1.6 xx 10^(-19)))=2.668 xx 10^(-9)` m
` lambda= 2.67 xx 10^(-9) m`
The two wavelength of the emitted electron is ` lt 2.75 xx 10^(-9)`m