The work function of a metal is w0 and l is wavelength of incident radiation. There is no emission of photoelectron whenThe work function of a metal is w0 and l is wavelength of incident radiation. There is no emission of photoelectron when\

 

According to Einstein’s photoelectric equation maximum kinetic energy of photoelectrons, $K{E}_{ max }=E_v-\phi$

or   $2=5-\phi$ $\therefore \phi =3eV$

When $E_v=6eV$  then, $K{E}_{ max }=6-3=3eV$

or $e\left(V_{cathode}-V_{anode}\right)=3eV$

or   $V_{cathode}-V_{anode}=3V=-V_{stopping}$

  $\therefore V_{stopping}=-3V$

Intersity a number of photons kinetic Energy a f

Incident energy $=2.20\mathrm{e}\mathrm{V}$

If $?<2.20\mathrm{e}\mathrm{V}$  electron will emit.

$?>2.20\mathrm{e}\mathrm{V}\mathrm{}$ No electron emission

Only caesium will emit electron

$eVs=hv-?$

$\begin{array}{rr}& \text{At}{\mathrm{V}}_{\mathrm{s}}=0\Rightarrow \mathrm{h}\mathrm{v}=?\\ & \Rightarrow \mathrm{}?=\left. [6.62\times {10}^{-34}\right]\left. [{10}^{14}\right]\left[5.5\right]\\ & \Rightarrow \mathrm{}?=\frac{\left. [6.62\times {10}^{-34}\right]\left. [{10}^{14}\right]\left[5.5\right]\mathrm{e}\mathrm{V}}{\left(1.6\times {10}^{-19}\right]}\end{array}$

$=2.27$

Relation between De-Broglie wavelength and K.E. is

$\lambda =\frac{\mathrm{h}}{\sqrt[]{2\left(\mathrm{K}\mathrm{E}\right){\mathrm{m}}_{\mathrm{e}}}}\Rightarrow \lambda \propto \frac{1}{\sqrt[]{\mathrm{K}\mathrm{E}}}$

$\begin{array}{rr}& \frac{{\lambda }_{\mathrm{A}}}{{\lambda }_{\mathrm{B}}}=\frac{\sqrt[]{{\mathrm{K}\mathrm{E}}_{\mathrm{B}}}}{\sqrt[]{\mathrm{K}\mathrm{E}}}\Rightarrow \frac{1}{2}=\sqrt[]{\frac{{\mathrm{T}}_{\mathrm{A}}-1.5}{{\mathrm{T}}_{\mathrm{A}}}}\\ & \Rightarrow {\mathrm{T}}_{\mathrm{A}}2\mathrm{e}\mathrm{V}\\ & \therefore {\mathrm{K}\mathrm{E}}_{\mathrm{B}}=2-1.5=0.5\mathrm{e}\mathrm{V}\\ & {?}_{\mathrm{B}}=4.5-0.5=4\mathrm{e}\mathrm{V}\end{array}$