The radiation corresponding to 3 → 2 transition of a hydrogen atom falls on a gold surface to generate…

$3\to 2\Rightarrow 1.89 \mathrm{eV}$
$5\times {10}^{-4} \mathrm{T} r=7 \mathrm{mm}$
$r=\frac{mv}{qB}\Rightarrow mv=qrB$
$\Rightarrow E=\frac{P^2}{2 m}=\frac{(qRB{)}^2}{2 m}$
$=\frac{{\left(1.6\times {10}^{-19}\times 7\times {10}^{-3}\times 5\times {10}^{-4}\right)}^2}{2\times 9.1\times {10}^{-31}\mathrm{Joule}}$
$=\frac{3136\times {10}^{-52}}{18.2\times {10}^{-31}\times 1.6\times {10}^{-19}} \mathrm{eV}$
$=1.077 \mathrm{eV}$
We know work function $=$ energy incident $-$ $(\mathrm{KE}{)}_{\text{electron }}$

$\varphi =1.89-1.077=0.813 \mathrm{eV}$