The radiation corresponding to 3 → 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of $5\times {10}^{-4}\mathrm{T}$. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work-function of the metal is (Take, mass of electron $=9.1\times {10}^{-31}\mathrm{kg}$)

Energy of photon can be given as

${\mathrm{E}}_{\mathrm{p}}=13.6\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]\mathrm{eV}$

where, n₁ = lower energy level

and n₂ = higher energy level.

As per question, n₁ = 2, n₂ = 3

$\begin{array}{c}\therefore {\mathrm{E}}_{\mathrm{p}}=13.6\left[\frac{1}{(2{)}^2}-\frac{1}{(3{)}^2}\right]\\ =13.6\left[\frac{1}{4}-\frac{1}{9}\right]=13.6\left[\frac{9-4}{36}\right]=1.89\mathrm{eV}\end{array}$

We know that work-function is the minimum energy required to eject photoelectrons from metal surface.

For gold plate, it will be

$\mathrm{\varphi }={\mathrm{E}}_{\mathrm{P}}-{\mathrm{KE}}_{max}$                           ...(i)

[Given, $\mathrm{B}=5\times {10}^{-4}\mathrm{T},\mathrm{r}=7\mathrm{mm}=7\times {10}^{-3}\mathrm{m}$, $\mathrm{q}=1.6\times {10}^{-19}\mathrm{C}\text{ and }\mathrm{m}=9.1\times {10}^{-31}\mathrm{kg}$]

Therefore, velocity of photoelectrons will be

$\mathrm{v}=\frac{\mathrm{Bqr}}{\mathrm{m}}=\frac{5\times {10}^{-4}\times 1.6\times {10}^{-19}\times 7\times {10}^{-3}}{9.1\times {10}^{-31}}=6.15\times {10}^5{\mathrm{ms}}^{-1}$

Kinetic energy will be

$\therefore \mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2=\frac{1\times 9.1\times {10}^{-31}\times {\left(6.15\times {10}^5\right)}^2}{2\times 1.6\times {10}^{-19}}\mathrm{eV}=1.075\mathrm{eV}$

Now, putting the values, in Eq. (i), we get

$\mathrm{\varphi }=(1.89-1.075)\mathrm{eV}=0.82\mathrm{eV}$