Two photons having energies twice and thrice the work function of a metal are incident one after another on the metal surface. Then the ratio of maximum velocities of the photoelectrons emitted in the two cases is respectively:

According to Einstein's photoelectric equation,the graph of the maximum kinetic energy of emitted photoelectrons versus the frequency of incident radiation is a straight line. Its slope . . . . . .

Radiation of monochromatic waves with a wavelength of $400 \ nm$ is incident on the surfaces of $Zn$,$Fe$,and $Ni$ metals,which have work functions of $3.4 \ eV$,$4.8 \ eV$,and $5.9 \ eV$ respectively. (Take $hc = 1242 \ eV \ nm$)
$(a)$ The maximum $KE$ of photoelectrons emitted from any metal surface is $0.3 \ eV$.
$(b)$ No photoelectrons are emitted from the surface of $Ni$.
$(c)$ If the frequency of the radiation source is doubled,the $KE$ of the photoelectrons also doubles.
$(d)$ If the wavelength of the incident radiation is less than $200 \ nm$,photoelectrons will be emitted from the surfaces of all three metals.
The correct statements are:

$A$ metal surface of work function $1.07 eV$ is irradiated with light of wavelength $332 nm$. The retarding potential required to stop the escape of photo-electrons is .............. $V$.

When a metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $2\lambda$,the stopping potential is $\frac{V}{4}$. The threshold wavelength for the metallic surface is:

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