In the case of photoelectric emission from a certain metal,the cutoff frequency is $v$. If radiation of frequency $2v$ is incident on the metal plate,the maximum possible velocity of the emitted electrons will be ($m=$ mass of electron).

When light of wavelength '$\lambda$' is incident on a photosensitive surface, the stopping potential is '$V$'. When a light of wavelength $1.5\lambda$ is incident on the same surface, the stopping potential is '$\frac{V}{4}$'. Threshold wavelength for the surface is

According to the photoelectric effect,the plot of kinetic energy of the emitted photo-electrons from a metal versus the frequency of the incident radiation gives a straight line whose slope

For a photosensitive material, the work function is $W_0$ and the stopping potential is $V$. What is the wavelength of the incident radiation? ($h$ = Planck's constant, $c$ = velocity of light, $e$ = electronic charge)

By the photoelectric effect,Einstein proved:

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? $(h = 6.62 \times 10^{-34} \text{ J s})$