A metal surface is illuminated by a radiation of wavelength 4500 ∘...

From the Einstein's photoelectric equation $K = \dfrac{hc}{\lambda} - ϕ ⇒ ϕ = \dfrac{hc}{\lambda} - K$

Now $R = \dfrac{mv}{Bq} ⇒ v = \dfrac{RBq}{m} ⇒ K = \dfrac{{R}^{2}{B}^{2}{q}^{2}}{2m}$

Therefore, $ϕ = \dfrac{hc}{\lambda} - \dfrac{{R}^{2}{B}^{2}{q}^{2}}{2m} = \dfrac{6 . 63 \times {10}^{- 34} \times 3 \times {10}^{8}}{4500 \times {10}^{- 10}} - \dfrac{{\left(2 \times {10}^{- 3} \times 2 \times {10}^{- 3} \times 1 . 6 \times {10}^{- 19}\right)}^{2}}{2 \times 9 . 1 \times {10}^{- 31}} ⇒ ϕ = \left(4 . 42 - 2 . 25\right) \times {10}^{- 19}J = \dfrac{2 . 27}{1 . 6}eV = 1 . 36eV$