Radiation, with wavelength $$6561$$ Å falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $$3 \times 10^{-4}$$ T. If the radius of the largest circular path followed by the electrons is $$10$$ mm, the work function of the metal is close to:

We begin with the photo-electric equation, which states that for each photon

$$h\nu \;=\; \dfrac{1}{2}m_ev_{\text{max}}^{\,2} \;+\; \phi,$$

where $$h$$ is Planck’s constant, $$\nu$$ is the frequency of the incident radiation, $$m_e$$ is the mass of the electron, $$v_{\text{max}}$$ is the maximum speed of the emitted electrons, and $$\phi$$ is the work function of the metal.

First we convert the given wavelength into metres. One angstrom equals $$10^{-10}\,\text{m}$$, so

$$\lambda \;=\; 6561~\text{Å} \;=\; 6561 \times 10^{-10}\,\text{m} \;=\; 6.561 \times 10^{-7}\,\text{m}.$$

Using the relation $$c = \lambda \nu,$$ we have

$$\nu = \dfrac{c}{\lambda} = \dfrac{3.00 \times 10^{8}\,\text{m s}^{-1}}{6.561 \times 10^{-7}\,\text{m}} = 4.573 \times 10^{14}\,\text{s}^{-1}.$$

Hence the photon energy is

$$h\nu = \left(6.626 \times 10^{-34}\,\text{J s}\right)\!\left(4.573 \times 10^{14}\,\text{s}^{-1}\right) = 3.032 \times 10^{-19}\,\text{J}.$$

To express this in electron-volts we divide by $$1\,\text{eV}=1.602\times10^{-19}\,\text{J}$$:

$$h\nu = \dfrac{3.032 \times 10^{-19}\,\text{J}}{1.602 \times 10^{-19}\,\text{J/eV}} = 1.895\,\text{eV}.$$

Next we use the information about the electron’s motion in the magnetic field. An electron entering a uniform field $$B$$ perpendicular to its velocity describes a circle whose radius $$r$$ satisfies the equality of magnetic and centripetal forces:

$$e v_{\text{max}} B = \dfrac{m_e v_{\text{max}}^{\,2}}{r}.$$

Solving for $$v_{\text{max}}$$ gives

$$v_{\text{max}} = \dfrac{e B r}{m_e}.$$

Substituting the numerical values $$e = 1.602 \times 10^{-19}\,\text{C},\; B = 3.0 \times 10^{-4}\,\text{T},\; r = 10\,\text{mm} = 0.010\,\text{m},\; m_e = 9.11 \times 10^{-31}\,\text{kg},$$ we get

$$v_{\text{max}} = \dfrac{(1.602 \times 10^{-19})(3.0 \times 10^{-4})(0.010)}{9.11 \times 10^{-31}} = \dfrac{4.806 \times 10^{-25}}{9.11 \times 10^{-31}} = 5.27 \times 10^{5}\,\text{m s}^{-1}.$$

Now we find the kinetic energy of these fastest electrons:

$$\dfrac{1}{2}m_e v_{\text{max}}^{\,2} = \dfrac{1}{2}(9.11 \times 10^{-31}\,\text{kg})\bigl(5.27 \times 10^{5}\,\text{m s}^{-1}\bigr)^{2}.$$

Calculating the square first, $$\bigl(5.27 \times 10^{5}\bigr)^{2} = 2.777 \times 10^{11},$$ so

$$\dfrac{1}{2}m_e v_{\text{max}}^{\,2} = 0.5 \times 9.11 \times 10^{-31} \times 2.777 \times 10^{11} = 1.265 \times 10^{-19}\,\text{J}.$$

Converting this to electron-volts:

$$\dfrac{1.265 \times 10^{-19}\,\text{J}}{1.602 \times 10^{-19}\,\text{J/eV}} = 0.790\,\text{eV}.$$

Finally, using the photo-electric equation $$h\nu = \dfrac{1}{2}m_ev_{\text{max}}^{\,2} + \phi,$$ we obtain the work function

$$\phi = h\nu - \dfrac{1}{2}m_ev_{\text{max}}^{\,2} = 1.895\,\text{eV} - 0.790\,\text{eV} = 1.105\,\text{eV}.$$

Rounding to the provided precision, the work function is about $$1.1\,\text{eV}.$$

Hence, the correct answer is Option C.