Photoelectric threshold wavelength for a metal is `3800 Å`. Find the maximum kinetic energy of emitted photoelectron, when ultraviolet radiation of wave length `2000Å` is incident on the metal surface. Planck's constant, `h = 6.62 xx 10^(-34) J.s`

Explanation

Maximum kinetic energy of photoelectron,
`E_("max") = hf - W_(0) = hf - hf_(0) = (hc)/(lamda) - (hc)/(lamda_(0))`
`= hc ((1)/(lamda) - (1)/(lamda_(0)))= hc (lamda_(0) - lamda)/(lamda lamda_(0))`
In this case, wavelength of incident light,
`lamda = 2000 Å = 2000 xx 10^(-10) m = 2 xx 10^(-7) m`
Threshold wavelength,
`lamda_(0) = 3800 xx 10^(-1) m = 3.8 xx 10^(-7)m`
Hence, `E_("max") = (6.62 xx 10^(-34)) xx (3 xx 10^(8)) xx ((3.8 -2) xx 10^(-7))/(3.8 xx 2 xx 10^(-14))J`
`= (6.62 xx 10^(-34) xx 3 xx 10^(8) xx 1.8)/(3.8 xx 2 xx 10^(-7) xx 1.6 xx 10^(-19)) eV`
= 2.94 eV