(i) The photoemission of electrons does not depend on the intensity but it depends on the frequency and hence on the energy of a photon of incident light.
(ii) If the energy of a photon is greater than the work function, the photoemission of electrons results however weak the incident radiation may be.
(iii) The energy of a photon of red light is less than the work function of zinc, so red light cannot emit photoelectrons.
(iv) The energy of a photon of ultraviolet light is greater than the work function of zinc, so ultraviolet light can emit photoelectrons.

\(\frac{\lambda}{2}\)
Reason: de Broglie wavelength associated with the electron is
\(\lambda =\frac { h }{ \sqrt { 2mqV } } \Rightarrow \lambda \alpha \frac { 1 }{ \sqrt { V } } \\ \)
Obviously, when accelerating potential becomes 4 V, the de- Broglie wavelength reduces to half.

Work function W = hv_o = \(\frac{hc}{\lambda_o}\)
\({ \lambda }_{ o }\alpha \frac { 1 }{ W } \)
As W_AB; (\({ \lambda }_{ o }\)) _A > (\({ \lambda }_{ o }\)) _B
i.e., the threshold wavelength of metal A is higher.

From Einstein's photoelectric equation, stopping potential V_o is
eVo = hv - hv_o V_o = \(\frac{h}{e}(v-v_o)\)
Given > v_o, so with increase of frequency v, stopping potential increases.

Curves a and b have different intensities but the same stopping potential, so curves 'a' and 'b' have the same frequency but different intensities.