Let threshold frequency of emitter plate \(=v_0\)
Energy of photon in first case is \(\mathrm{E}\).
When switch \(S_1\) is closed and switch \(S_2\) is open,
So, \(E=h v_0+(5+1) \mathrm{eV}\) ...(i)
For second case, when switch \(S_1\) is open and switch \(S_2\) is closed and frequency of incident light is doubled.
then, \(\quad 2 E=h v_0+(20-5) \mathrm{eV}\) ...(ii)
From Eqs. (i) and (ii), we get
\(\begin{aligned}
& \Rightarrow \quad 2\left(h v_0+6 \mathrm{eV}\right)=h v_0+15 \mathrm{eV} \\
& \Rightarrow \quad 2 h v_0+12 \mathrm{eV}=h v_0+15 \mathrm{eV} \\
& \Rightarrow \quad h v_0=3 \mathrm{eV} \\
& \Rightarrow \quad v_0=\frac{3 \times 1.6 \times 10^{-19}}{6.62 \times 10^{-34}}=7.25 \times 10^{14} \mathrm{~Hz} \\
& \lambda_0=\frac{c}{v_0}=\frac{3 \times 10^8}{7.25 \times 10^{14}} \\
& \lambda_0=41333 Å \\
\end{aligned}\)
Hence, the threshold wavelength of the emitter plate is \(4133.3 Å\).