The threshold wavelength for photoelectric emission in tungsten is $400 \ nm$. The wavelength of light that must be used in order to eject electrons with a maximum kinetic energy of $0.9 \ eV$ is .............. $nm$.

In the photoelectric effect,if the intensity of light is doubled,then the maximum kinetic energy of the photoelectrons will become:

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:

In a photoelectric experiment,the stopping potential was measured to be $V_1$ and $V_2$ volts with incident light of wavelength $\lambda$ and $\frac{\lambda}{2}$ respectively. The value of $V_2$ is: [where $\phi=$ work function,$e=$ electronic charge]

When photons of energy $8 \times 10^{-19} \ J$ are incident on a photosensitive material,the de Broglie wavelength of the photoelectrons emitted with maximum kinetic energy is $10 \ Å$. The work function of the photosensitive material is nearly (in $eV$)

When radiation of wavelength $\lambda$ is used to illuminate a metallic surface,the stopping potential is $V.$ When the same surface is illuminated with radiation of wavelength $3 \lambda,$ the stopping potential is $\frac{V}{4}.$ If the threshold wavelength for the metallic surface is $n \lambda,$ then the value of $n$ will be......