The radiation corresponding to $$3 \to 2$$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of $$3 \times 10^{-4}$$ T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to:

For a hydrogen atom, the energy of an electron in the $$n^{\text{th}}$$ orbit is given by the well-known Bohr expression $$E_n=-\dfrac{13.6\ \text{eV}}{n^2}$$. When the electron jumps from a higher level $$n_1$$ to a lower level $$n_2$$, the photon released carries the energy difference

$$E_{\text{photon}} = 13.6\ \text{eV}\left(\dfrac{1}{n_2^2}-\dfrac{1}{n_1^2}\right).$$

Here the transition is $$3 \rightarrow 2$$, so we have

$$E_{\text{photon}} = 13.6\ \text{eV}\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right) = 13.6\ \text{eV}\left(\dfrac14-\dfrac19\right) = 13.6\ \text{eV}\left(\dfrac{9-4}{36}\right) = 13.6\ \text{eV}\left(\dfrac{5}{36}\right) = \dfrac{68}{36}\ \text{eV} = 1.889\ \text{eV}\;(\text{approximately}).$$

This photon strikes a metal surface and liberates an electron. According to Einstein’s photoelectric equation,

$$E_{\text{photon}} = \phi + K_{\text{max}},$$

where $$\phi$$ is the work function of the metal and $$K_{\text{max}}$$ is the maximum kinetic energy of the emitted electron. We therefore need $$K_{\text{max}}$$.

The liberated electrons enter a uniform magnetic field $$B = 3.0 \times 10^{-4}\ \text{T}$$ perpendicular to their velocity, describing circular paths. For a charge $$e$$ of mass $$m_e$$ moving with speed $$v$$ in a perpendicular magnetic field, the magnetic (Lorentz) force provides the centripetal force:

$$e v B = \dfrac{m_e v^2}{r} \;\;\Rightarrow\;\; v = \dfrac{e B r}{m_e}.$$

The given radius of the largest orbit is $$r = 10.0\ \text{mm} = 0.010\ \text{m}$$. Substituting the electronic charge $$e = 1.602 \times 10^{-19}\ \text{C}$$, the electron mass $$m_e = 9.11 \times 10^{-31}\ \text{kg}$$, the field $$B = 3.0 \times 10^{-4}\ \text{T}$$ and the radius into the expression for speed, we could find $$v$$, but it is more direct to calculate the kinetic energy right away. The kinetic energy is

$$K_{\text{max}} = \dfrac12 m_e v^2 = \dfrac12 m_e \left(\dfrac{e B r}{m_e}\right)^2 = \dfrac{e^2 B^2 r^2}{2m_e}.$$

Putting in the numbers step by step:

$$e^2 = (1.602 \times 10^{-19})^2 = 2.566 \times 10^{-38},$$

$$B^2 = (3.0 \times 10^{-4})^2 = 9.0 \times 10^{-8},$$

$$r^2 = (0.010)^2 = 1.0 \times 10^{-4},$$

Multiplying these three factors,

$$e^2 B^2 r^2 = 2.566 \times 10^{-38}\; \times 9.0 \times 10^{-8}\; \times 1.0 \times 10^{-4} = 2.309 \times 10^{-49}\ \text{J}^2\!\cdot\!\text{s}^2/\text{kg},$$

and dividing by $$2m_e$$,

$$2m_e = 2 \times 9.11 \times 10^{-31} = 1.822 \times 10^{-30},$$

we get

$$K_{\text{max}} = \dfrac{2.309 \times 10^{-49}}{1.822 \times 10^{-30}} = 1.268 \times 10^{-19}\ \text{J}.$$

To express this in electron-volts, we recall $$1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$$, so

$$K_{\text{max}} = \dfrac{1.268 \times 10^{-19}}{1.602 \times 10^{-19}}\ \text{eV} \approx 0.79\ \text{eV}.$$

Finally we return to Einstein’s equation:

$$\phi = E_{\text{photon}} - K_{\text{max}} = 1.889\ \text{eV} - 0.79\ \text{eV} \approx 1.10\ \text{eV}.$$

Hence, the correct answer is Option B.