For a certain metal, when monochromatic light of wavelength \(\lambda\) is incident, the stopping potential for photoelectrons is \(3V_0\). When the same metal is illuminated by light of wavelength \(2\lambda\), then the stopping potential becomes \(V_0\). The threshold wavelength for photoelectric emission for the given metal is \(\alpha\lambda\). The value of \(\alpha\) is _______.

Using Einstein's photoelectric equation:

\(eV_s = \dfrac{hc}{\lambda} - \Phi\)

For the first case:

\(3eV_0 = \dfrac{hc}{\lambda} - \Phi\)

For the second case:

\(eV_0 = \dfrac{hc}{2\lambda} - \Phi\)

Multiplying the second equation by \(3\) and equating with the first equation:

\(\dfrac{hc}{\lambda} - \Phi = \dfrac{3hc}{2\lambda} - 3\Phi\)

\(2\Phi = \dfrac{3hc}{2\lambda} - \dfrac{hc}{\lambda}\)

\(2\Phi = \dfrac{hc}{2\lambda}\)

\(\Phi = \dfrac{hc}{4\lambda}\)

Since the work function \(\Phi = \dfrac{hc}{\lambda_0}\), where \(\lambda_0\) is the threshold wavelength:

\(\dfrac{hc}{\lambda_0} = \dfrac{hc}{4\lambda}\)

\(\lambda_0 = 4\lambda\)

Given \(\lambda_0 = \alpha\lambda\), we get \(\alpha = 4\).

Answer: \(4\)