Radiation of wavelength $\lambda$ is incident on a photocell. The fastest emitted electron has speed $v$. If the wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be:

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

The work function of a metallic surface is $5.01 \, eV$. The photo-electrons are emitted when light of wavelength $2000 \, \mathring{A}$ falls on it. The potential difference applied to stop the fastest photo-electrons is ............... $volt$ $[h = 4.14 \times 10^{-15} \, eV \cdot s]$

Which one of the following graphs correctly represents the variation of maximum kinetic energy $(E_{k})$ of the emitted electrons with frequency $(\nu)$ of incident light in the photoelectric effect?

When light of frequency $v$ is incident on two metallic plates $A$ and $B$,photoelectrons are emitted. If the work function of $A$ is greater than that of $B$,which of the following curves correctly represents the relationship between the stopping potential $V$ and the incident frequency $v$?

Statement $-1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by $X$-rays, both $V_0$ and $K_{max}$ increase.
Statement $-2$: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.