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

Answer

A

Explanation

According the Einsteni.s photoelectric equation , the maximum kinetic energy of the emitted electron is
`K_(max) = (hc)/lamda - phi_0`
where `lamda` is the wavelength of incident light and `phi_0` is the work function .
Here, `lamda = 500 nm, 1240 eV nm and phi_0 =2.28 eV`
`:. K_(max) = (1240 eV nm) /(500 nm) - 2.28 eV`
The de Broglie wavelength of the emitted electron is
`lamda_(min) = h/(sqrt(2mK_(max))`
where h is Planck.s constant and m is the mass of the electron.
As `h = 6.6 xx 10^(-34) Js , m = 9 xx 10^(-31) kg`
and `K_(max) = 0.2 eV = 0.2 xx 1.6 xx 10^(-19) J`
`:. lamda_(min) = (6.6 xx10^(-34)Js)/(sqrt(2(9 xx 10^(-31) kg)(0.2 xx 1.6 xx 10^(-19)J)))`
`= (6.6)/(2.4) xx 10^(-9) m = 2.8 xx 10^(-9) m`
So, `lamda ge 2.8 xx 10^(-9)m`