When light falls on a metal surface,the maximum kinetic energy of the emitted photo-electrons depends upon

Light of wavelength $\lambda$ falls on a metal having work function $\frac{hc}{\lambda_0}$. Photoelectric effect will take place only if ($\lambda_0$ is the threshold wavelength).

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electric potential to the photoemissive substance.
Reason $(R) :$ $A$ negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation.
In the light of the above statements, choose the most appropriate answer from the options given below:

The electric field component of an $EM$ radiation varies with time as $E = a(\cos \omega_{0} t + \sin \omega t \cos \omega_{0} t)$, where '$a$' is a constant, $\omega = 10^{15} \text{ s}^{-1}$, and $\omega_{0} = 5 \times 10^{15} \text{ s}^{-1}$. This radiation falls on a metal whose stopping potential is $2 \text{ V}$. Which of the following statement$(s)$ is/are true? $(h = 6.62 \times 10^{-34} \text{ J s})$

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential of ejected photoelectrons is $4.8 \, V$. If the same surface is illuminated by radiation of double the previous wavelength,then the stopping potential becomes $1.6 \, V$. The threshold wavelength of the metal is $... \lambda$.

The work function of caesium is $2.14\, eV$. Find the wavelength of the incident light if the photo current is brought to zero by a stopping potential of $0.60\, V$. (Result in $nm$)