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Step-by-Step Solution
Step 1: Identify the relevant photoelectric equations
For a light of wavelength \lambda , the maximum kinetic energy K_{\text{max}} of the emitted electrons is given by
K_{\text{max}} = \dfrac{hc}{\lambda} - \phi,
where
h is Planck’s constant,
c is the speed of light, and
\phi is the work function of the metal.
The stopping potential V_s is related to the maximum kinetic energy by
e \, V_s = K_{\text{max}},
where e is the charge of the electron.
Step 2: Write down the equations given in the problem
1) When radiation of wavelength \lambda is used, the stopping potential is 3V_0 :
e \,(3V_0) = \dfrac{hc}{\lambda} - \phi \quad \dots (1)
2) When radiation of wavelength 2 \lambda is used, the stopping potential is V_0 :
e \,(V_0) = \dfrac{hc}{2 \lambda} - \phi \quad \dots (2)
Step 3: Subtract the second equation from the first to eliminate \phi
Subtract equation (2) from equation (1):
\Bigl[\dfrac{hc}{\lambda} - \phi\Bigr] - \Bigl[\dfrac{hc}{2\lambda} - \phi\Bigr] = e(3V_0) - e(V_0).
Notice that -\,\phi and +\, \phi cancel out, giving:
\dfrac{hc}{\lambda} - \dfrac{hc}{2 \lambda} = e(3V_0 - V_0).
Simplify both sides:
\dfrac{hc}{2\lambda} = 2 e V_0
\quad \Longrightarrow \quad
\dfrac{hc}{\lambda} = 4 e V_0.
Step 4: Solve for the work function \phi
Use either of the original equations to find \phi . Substituting \dfrac{hc}{\lambda} = 4 e V_0 into equation (1), for instance:
e (3V_0) = 4 e V_0 - \phi
\quad \Longrightarrow \quad
\phi = 4 e V_0 - 3 e V_0 = e V_0.
Hence,
\phi = e V_0.
Step 5: Find the threshold wavelength \lambda_t
The threshold wavelength \lambda_t is defined by the condition that the maximum kinetic energy becomes zero. In other words,
\phi = \dfrac{hc}{\lambda_t}.
Since \phi = e V_0 , we have
e V_0 = \dfrac{hc}{\lambda_t}
\quad \Longrightarrow \quad
\lambda_t = \dfrac{hc}{e V_0}.
From Step 3, we also know
\dfrac{hc}{\lambda} = 4 e V_0,
so
e V_0 = \dfrac{hc}{4\lambda}.
Therefore,
\lambda_t = \dfrac{hc}{e V_0}
= \dfrac{hc}{\tfrac{hc}{4 \lambda}}
= 4 \lambda.
Final Answer
The threshold wavelength of the surface is
4 \lambda.
