Solution: Threshold Wavelength with Retarding Field
Key Concept: The Work-Energy Theorem. The maximum kinetic energy of the emitted photoelectron ($K_{max}$) is entirely consumed by the work done against the retarding electric field ($W = Fd$) to stop it at distance $d$.
Step 1: Energy Balance
The electron stops after traveling distance $d$ against field $E$. The work done by the field is $eEd$. $$ K_{max} = eEd $$
The electron stops after traveling distance $d$ against field $E$. The work done by the field is $eEd$. $$ K_{max} = eEd $$
Step 2: Photoelectric Equation
Using Einstein’s equation: $$ K_{max} = \frac{hc}{\lambda} – \phi $$ Where $\phi = \frac{hc}{\lambda_0}$ is the work function. $$ eEd = \frac{hc}{\lambda} – \frac{hc}{\lambda_0} $$
Using Einstein’s equation: $$ K_{max} = \frac{hc}{\lambda} – \phi $$ Where $\phi = \frac{hc}{\lambda_0}$ is the work function. $$ eEd = \frac{hc}{\lambda} – \frac{hc}{\lambda_0} $$
Step 3: Solve for $\lambda_0$
Rearrange the terms: $$ \frac{hc}{\lambda_0} = \frac{hc}{\lambda} – eEd $$ Divide by $hc$: $$ \frac{1}{\lambda_0} = \frac{1}{\lambda} – \frac{eEd}{hc} $$ Invert both sides: $$ \lambda_0 = \left( \frac{1}{\lambda} – \frac{eEd}{hc} \right)^{-1} $$
Rearrange the terms: $$ \frac{hc}{\lambda_0} = \frac{hc}{\lambda} – eEd $$ Divide by $hc$: $$ \frac{1}{\lambda_0} = \frac{1}{\lambda} – \frac{eEd}{hc} $$ Invert both sides: $$ \lambda_0 = \left( \frac{1}{\lambda} – \frac{eEd}{hc} \right)^{-1} $$
Correct Option: (d) $\left( \frac{1}{\lambda} – \frac{eEd}{hc} \right)^{-1}$