The radiation corresponding to 3 → 2 transition of a hydrogen atom...

$3 \to 2 ⇒ 1 . 89eV$
$5 \times {10}^{- 4}Tr = 7mm$
$r = \dfrac{mv}{qB} ⇒ mv = qrB$
$⇒ E = \dfrac{{P}^{2}}{2m} = \dfrac{( qRB{)}^{2}}{2m}$
$= \dfrac{{\left(1 . 6 \times {10}^{- 19} \times 7 \times {10}^{- 3} \times 5 \times {10}^{- 4}\right)}^{2}}{2 \times 9 . 1 \times {10}^{- 31}Joule}$
$= \dfrac{3136 \times {10}^{- 52}}{18 . 2 \times {10}^{- 31} \times 1 . 6 \times {10}^{- 19}}eV$
$= 1 . 077eV$
We know work function = energy incident $-$ $( KE{)}_{\text{electron}}$

$ϕ = 1 . 89 - 1 . 077 = 0 . 813eV$