$A$ photon of energy $hv$ is absorbed by a free electron of a metal having work function $\phi < hv$.

Which theory of light can explain the photoelectric effect?

The work functions for metals $A$, $B$, and $C$ are $1.92 eV$, $2.0 eV$, and $5 eV$ respectively. The metal(s) which will emit photoelectrons for incident radiation of wavelength $4100 Å$ is/are $[h=6.63 \times 10^{-34} J s, e=1.6 \times 10^{-19} C, c=3 \times 10^8 m/s]$.

If in a photoelectric cell,the wavelength of incident light is changed from $4000 \mathring A$ to $3000 \mathring A$,then the change in stopping potential will be ..... $V$.

The electric field associated with a monochromatic light wave is given by $E = E_0 \sin \left[ \left( 1.57 \times 10^7 \text{ m}^{-1} \right) (x - ct) \right]$. The stopping potential when this light is used in a photoelectric experiment with a metal having a work function of $1.9 \text{ eV}$ is: (Planck's constant,$h = 6.64 \times 10^{-34} \text{ J-s}$) (in $\text{ V}$)

The variation of stopping potential for metals $A$,$B$,$C$ and $D$ with the frequency of incident radiation is shown in the figure. For which metal is the stopping potential higher for a given frequency of incident radiation $(v)$ if the threshold frequency is lower $(
u_o)$?