QJEEAdv15Ph08a2: JEE Advanced 2015 Physics - Class 12 JEEAdv15Ph08a2 Solution: For photo-electric effect with incident photon wavelength , the …

Stopping potential (V₀) is given by

$e{V_0} = {{hc} \over \lambda } - \phi$

Graph between V₀ and $\lambda$ :

$e{V_0} + \phi = {{hc} \over \lambda }$

$(e{V_0} + \phi )\lambda = hc$

$(e{V_0} + \phi )\lambda$ = constant

Here, both e and $\phi$ are also constant. It represents a hyperbola.

For, V₀ = 0, $\lambda = {{constant} \over \phi } =$ constant

So correct option is (a).

Graph between V₀ and ${1 \over \lambda }$ :

${V_0} = \left( {{{hc} \over e}} \right)\left( {{1 \over \lambda }} \right) - \left( {{\phi \over e}} \right)$

It represents a straight line with slope $\left( {{{hc} \over e}} \right)$ and intercept $\left( { - {\phi \over e}} \right)$ on V₀ axis.