If a light of wavelength 330 nm is incident on a metal with work function 3.55 eV, the electrons are emitted. Then the wavelength of the emitted electron is (Take 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(2 m KE)) = (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`