Threshold Wavelength in Photoelectric Effect for Varying Incident Light

According to Einstein's photoelectric equation, the relationship between the stopping potential $V_s$, the incident wavelength $\lambda$, and the work function $\phi$ of a metal is given by: $e V_s = \frac{hc}{\lambda} - \phi$

Let $\lambda_{th} = \alpha\lambda$ be the threshold wavelength for the metal, so the work function can be expressed as: $\phi = \frac{hc}{\lambda_{th}} = \frac{hc}{\alpha\lambda}$

Case 1: When light of wavelength $\lambda$ is incident, the stopping potential is $3V_0$: $3e V_0 = \frac{hc}{\lambda} - \frac{hc}{\alpha\lambda} \quad \text{--- (1)}$

Case 2: When light of wavelength $2\lambda$ is incident, the stopping potential is $V_0$: $e V_0 = \frac{hc}{2\lambda} - \frac{hc}{\alpha\lambda} \quad \text{--- (2)}$

Multiplying equation (2) by $3$: $3e V_0 = \frac{3hc}{2\lambda} - \frac{3hc}{\alpha\lambda} \quad \text{--- (3)}$

Equating equations (1) and (3): $\frac{hc}{\lambda} - \frac{hc}{\alpha\lambda} = \frac{3hc}{2\lambda} - \frac{3hc}{\alpha\lambda}$

Dividing the entire equation by $\frac{hc}{\lambda}$: $1 - \frac{1}{\alpha} = \frac{3}{2} - \frac{3}{\alpha}$

Rearranging the terms to solve for $\alpha$: $\frac{3}{\alpha} - \frac{1}{\alpha} = \frac{3}{2} - 1$

$\frac{2}{\alpha} = \frac{1}{2}$

$\alpha = 4$

Thus, the threshold wavelength is $4\lambda$, and the value of $\alpha$ is $4$.