If the surface of a metal is successively exposed to radiation of wavelengths $\lambda_1 = 350 \, nm$ and $\lambda_2 = 450 \, nm$,the maximum velocity of photoelectrons differs by a factor of $2$. The work function of this metal is:

When a point source of light is at a distance of $50 \, cm$ from a photoelectric cell,the cut-off voltage is found to be $V_0$. If the same source is placed at a distance of $1 \, m$ from the cell,then the cut-off voltage will be

In a photoemissive cell with exciting wavelength $\lambda$,the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be

Assertion : Photosensitivity of a metal is high if its work function is small.
Reason : Work function $= hf_0$ where $f_0$ is the threshold frequency.

Light source having wavelength $331 \text{ nm}$ is used to generate photo-electrons whose stopping potential is $0.2 \text{ V}$. The work function of the used metal in the experiment is $\alpha \times 10^{-19} \text{ J}$. The value of $\alpha$ is . . . . . . . ($h = 6.62 \times 10^{-34} \text{ J s}$, $e = 1.6 \times 10^{-19} \text{ C}$ and $c = 3 \times 10^8 \text{ m/s}$) (in $.68$)

In the experiment of the photoelectric effect, if the frequency of incident light is $v_1$, the maximum kinetic energy $(K.E.)$ of photoelectrons is $K_0$. If the frequency of light is $v_2$, the maximum $K.E.$ of photoelectrons is $2K_0$. Which of the following relations is correct?