The critical wavelength for producing photoelectric effect in a metal is `2500 Å` What wavelength would be nuccesary be produce photoelectric effect from this metal , having twice the KE of these produced at `2000 Å`

Explanation

`KE_(2000) = hv_(2000) - hv_(0)= hc ((1)/(lambda_(2000)) - (1)/(lambda_(0)))`
`KE_("new") - hv_("new") = hv_(0)= hc ((1)/(lambda_("new")) - (1)/(lambda_(0)))`
KE_("new") = 2 xx KE_(2000)`
`2= (hc((1)/(lambda_("new")) - (1)/(2500)))/(hc((1)/(2000) - (1)/(2500))) , lambda = 1666.66 Å`