The threshold wavelength for emitting photons from a metal is `6.0 xx 10^3 Å`. What would be the wavelength of radiation to produce photoelectrons having twice the kinetic energy of those produced by radiation of wavelength `3 xx 10^3 Å` ?

Explanation

K.E. of photoelectrons emission is :
K.E. = `hv - hv_0 = (hc)/(lambda) - (hc)/(lambda_0)`
or `" " K.E. = hc (1/lambda - 1/(lambda_0))`
K.E. of electrons emitted by using `lambda = 3 xx 10^3 Å " or " 3000 Å`
`(K.E)_(1) = hc(1)/(3000 - 1/6000) ( :. lambda_0 = 6000 Å)`
K.E. of electrons by using wavelength `lambda` (to be calculated)
K.E. of electrons by using wavelength `lambda` (to be calculated)
`(K.E.)_(2) = hc (1/(lambda) - 1/(6000)`
Now, `(K.E.)_(2) = 2 (K.E.)_(1)`
`hc (1/(lambda) - 1/(6000)) = 2hc (1)/(3000 - 1/6000)`
or `1/lambda = 2/3000 - 1/(3000) + 1/6000 = 1/(2000)`
`:. lambda = 2000 Å`