When radiation of wavelength `4940 Å` is incident on a metal surface photoelectricity is generated. For a potential difference 0.6 V between cathode and anode, photocurrent stops. For another incident radiation, the stopping potential changes to 1.1V. Find work function of the metal and wavelength of the second radiation. `(h = 6.6 xx 10^(-27) " erg".s, e = 1.6 xx 10^(-19) C)`

Explanation

For the first radiation, stopping polential `V_(0) = 0.6V`
`:.` Maxnum kinetic energy of photoelectron,
`E_("max") = eV_(0) = 0.6eV`
Wavelength, `lamda = 4940 Å = 4940 xx 10^(-8) cm`
`:.` Energy of incident photon
`= hf = h(c)/(lamda) = (6.6 xx 10^(-27) xx 3 xx 10^(10))/(4940 xx 10^(-8))`
`= (6.6 xx 10^(-27) xx3 xx 10^(10))/(4940 xx 10^(-8) xx 1.6 xx 10^(-12)) eV = 2.5 eV`
If work function of the metal is `W_(0)`, from Einstein's equation,
`E_("max") = hf - W_(0)`
or, `W_(0) = hf - E_("max") = 2.5 - 0.6 = 1.9 eV`
For the second radiation, `V'_(0) = 1.1V`
Hence, `E'_("max") = 1.1eV`
`:. E'_("max") = hf' - W_(0)`
or, `hf' = E'_("max") + W_(0) = 1.1 + 1.9 = 3.0eV`
Hence, `(hf)/(hf') = (2.5)/(3.0) or, (f)/(f') = (5)/(6)`
or, `(c//lamda)/(c//lamda') = (5)/(6) or, (lamda')/(lamda) = (5)/(6)`
or, `lamda' = lamda xx (5)/(6) = 4940 xx (5)/(6) = 4117 Å`