Question 56: Dual Nature of Matter and Radiation

Question

For photo-electric effect with incident photon wavelength $$\lambda$$, the stopping potential is V0. Identify the correct variation(s) of V0 with $$\lambda$$ and $${1 \over \lambda }$$.

Accepted Answer

Stopping potential (V
0
) is given by
$$e{V_0} = {{hc} \over \lambda } - \phi $$
Graph between V
0
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, V
0
= 0, $$\lambda  = {{constant} \over \phi } = $$ constant
So correct option is (a).
Graph between V
0
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 V
0
axis.

Answer

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Dual Nature of Matter and Radiation

For photo-electric effect with incident photon wavelength $\lambda, the stopping potential is V0. Identify the correct variation(s) of V0 with \lambda and {1 \over \lambda }$.

Choose the correct answer: