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. `ge2.8 xx 10^(-9)m`
B. `le2.8 xx 10^(-12)m`
C. `lt2.8 xx 10^(-10)m`
D. `lt2.8xx10^(-9)m`

Answer

1

Explanation

According to Eistein.s photoelectric equation , the maximum kinetic energy of the emitted photoelectrons in the first case is
`K_(max1)=(hc)/(lambda)-phi_(0)`
`therefore K_(max1) = (1240eV nm)/(500 nm) - 2.28eV`
2.48 eV - 2,28 eV = 0.2eV
The de Broglie wavelenght of the emitted electron is
`lambda_(min) = (h)/(sqrt(2mK_(max)))`