Solution:
Solution
Let threshold wavelength be $$ \lambda_0 $$
Therefore work function for the surface will be $$ W= \dfrac{hc}{\lambda_0} $$
Stopping potential $$ V=\dfrac{E-W}{e} $$ where $$ E $$ is the energy of the incident beam ($$ \dfrac{hc}{\lambda} $$) and $$ e $$ is the charge on an electron.
Given
$$ 3eV_0=\dfrac{hc}{\lambda}-\dfrac{hc}{\lambda_0} $$
$$ eV_0=\dfrac{hc}{2\lambda}-\dfrac{hc}{\lambda_0} $$
Multiplying the second eqaution by 3 and subtracting it from the first we get
$$ \dfrac{2hc}{\lambda_0}-\dfrac{hc}{2\lambda}=0 $$
Solving we get $$ \lambda_0=4\lambda $$