When a piece of metal is illuminated by monochromatic light of wavelength '$\lambda$', the stopping potential for photoelectric current is $2.5 V_0$. When the same surface is illuminated by light of wavelength $1.5 \lambda$, the stopping potential becomes '$V_0$'. The value of threshold wavelength for photoelectric emission is:

$2.25 \lambda$

The correct answer is Option (4) → $2.25 \lambda$

$hv=\frac{1}{2}mv^2_{max}+W_0$ [Photoelectric equation] ...(1)

where,

$v$ → frequency

$v_{max}$ → Maximum velocity

$W_0$ → Work function

and,

$V_s$ (Stopping Potential) = $eV_s=K.E_{max}$  ...(2)

∴ from (1) and (2)

$hv=eV_s+W_0$

$⇒eV_s=hv-W_0$

and,

$V_s=2.5V_0$ when wavelenght = λ

$V_s=V_0$ when wavelength = 1.5λ

$2.5V_0×e=\frac{hc}{λ}-\frac{hc}{λ_0}$  ...(3)

$eV_0=\frac{hc}{1.5λ}-\frac{hc}{λ_0}$  ...(4)

∴ from (3) and (4) we get,

$2.5\frac{hc}{1.5λ}-2.5\frac{hc}{λ_0}=\frac{hc}{λ}-\frac{hc}{λ_0}$

$\frac{hc}{λ}\left(\frac{5}{3}-1\right)=\frac{hc}{λ_0}(1.5)$

$\frac{hc}{λ}×\frac{2}{3}=\frac{hc}{λ_0}×\frac{3}{2}$

$λ_0=\frac{3×3}{2×2}λ$

$=2.25λ$