Solution

Given :
The fringe width is $$y=1.0\times 2\ mm = 2.0 mm$$
The distance between the slits is $$ d = 0.24 mm$$
The work function of the material is $$W_0=2.2eV$$
The distance of the screen from the slits is $$D=1.2\ m$$

The fringe width is given as:
$$y=\dfrac { \lambda D }{ d } $$
or, $$\lambda =\dfrac { yd }{ D } =\dfrac { 2\times { 10 }^{ -3 }\times 0.24\times 10^{ -3 } }{ 1.2 } =4\times 10^{ -7 }m$$

The energy of the photon is obtained as
:
$$E=\dfrac { hc }{ \lambda } =\dfrac { 4.14\times 10^{ -15 }\times 3\times { 10 }^{ 8 } }{ 4\times 10 } =3.105\quad eV$$

Stopping potential $$eV_0=3.105-2.2=0.905V$$