The photoelectric threshold wavelength of certain material is `6000Å`. Determine the maximum energy in electron volts of the ejected photo electrons if the incident radiations has wavelength of `4000Å`.
(Given, `h=6.62xx10^(-34)Jsandc=3xx10^(8) ms^(-1)`)

Explanation

As, `KE_("max")=1/(2)mv_("max")^(2)=hv-hv_(0)=hc(1/(lamda)-1/(lamda_(0)))`
Here, `KE_("max")=6.62xx10^(-34)xx3xx10^(8)(1/(4000xx10^(-10))-1/(6000xx10^(-10)))`
`=6.62xx3xx1/(12)xx10^(-19)=1.655xx10^(-19)J`
`=(1.655xx10^(-19))/(1.6xx10^(-19))eV`
`therefore KE_("max")=1.034 eV`