When a certain metallic surface is illuminated with monochromatic light of wavelength `lamda`, the stopping potential for photoelectric current is `3V_0` and when the same surface is illuminated with light of wavelength `2lamda`, the stopping potential is `V_0`. The threshold wavelength of this surface for photoelectrice effect is

Options

A. `4 lamda`
B. `3.5 lamda`
C. `3 lamda`
D. `2.75 lamda`

Answer

A

Explanation

As we know , `hv = hv_0 + K.E_(max)`
`(hc)/(lamda_("incident")) = (hc)/(lamda_0) + eV_s`
when `lamda_("incident") = lamda, V_s = 3V`
and for `lamda_("incident") = 2 lamda , V_5 = 1V`
` therefore (hc)/(lamda) = (hc)/(lamda_0) = 3eV` ...(i)
and `(hc)/(2lamda) = (hc)/(lamda_0) + 1eV` ....(ii)
On simplifying (i) and (ii)
`(2hc)/(lamda_0) = 1/2 (hc)/(lamda) rArr lamda_0 = 4 lamda`