Radiation, with wavelength $6561 A falls on a metal surface to produce photoelectrons.

(1) Using Einstein's photoelectric equation,

$E=\omega_{0}+K E_{\max }$

$\Rightarrow \omega_{0}=K E_{\max }-E$

$p=\sqrt{2 m K E} \Rightarrow K E=\frac{p^{2}}{2 m}$

$r=\frac{p}{e B} \Rightarrow p=r e B$

$K_{\max }=\frac{r^{2} e^{2} B^{2}}{2 m}$

$K E_{\max }=\frac{12420}{\lambda}-\omega_{0}$

$\Rightarrow \omega_{0}=\frac{12420}{6561}-\frac{r^{2} e B^{2}}{2 m}($ In $\mathrm{VV})$

$=1.89(\mathrm{eV})-\frac{\left(10^{-4}\right)\left(1.6 \times 10^{-19}\right) 9 \times 10^{5}}{2 \times 9.07 \times 10^{-31}}$

$=(1.89-0.79) \mathrm{eV}=1.1 \mathrm{eV}$