Use Einstein’s photoelectric equation

$h\nu = \phi + K_{\max}$

where:

• $\phi = 2.14\,\text{eV}$ is the work function
• $K_{\max} = eV_0 = 2\,\text{eV}$ since stopping potential $V_0 = 2\,\text{V}$

Find the photon energy

$E = h\nu = 2.14 + 2 = 4.14\,\text{eV}$

Use the relation between energy and wavelength

$E = \frac{hc}{\lambda}$

So,

$\lambda = \frac{hc}{E} = \frac{1242\,\text{eV nm}}{4.14\,\text{eV}}$

Calculate

$\lambda = 300\,\text{nm}$

Match with the options

• A: $400\,\text{nm}$
• B: $600\,\text{nm}$
• C: $300\,\text{nm}$
• D: $200\,\text{nm}$

Hence, the correct option is C.