If in a photoelectric experiment,the wavelength of incident radiation is reduced from $6000 \, \mathring{A}$ to $4000 \, \mathring{A}$,then:

The electric field of a light wave is given as $\vec E = 10^{-3} \cos \left( \frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14} t \right) \hat x \, N/C$. This light falls on a metal plate with a work function of $2 \, eV$. The stopping potential of the photoelectrons is ................ $V$.

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then the threshold wavelength for the surface is

When a light of wavelength '$\lambda$' falls on the emitter of a photosensitive surface, the maximum speed of emitted photoelectrons is '$V$'. If the incident wavelength is changed to '$2\lambda/3$',the maximum speed of emitted photoelectrons will be

The correct graph between the maximum kinetic energy $(K_{\max})$ of a photoelectron and the inverse of the wavelength $(1/\lambda)$ of the incident radiation is given by the curve:

If two metals $A$ and $B$ are exposed to radiation of wavelength $350 \, nm$. The work functions of metals $A$ and $B$ are $4.8 \, eV$ and $2.2 \, eV$ respectively. Choose the correct option.