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Step-by-Step Solution
Step 1: Identify the Photon Energy from the Hydrogen Transition
The photon is emitted due to the transition of an electron in the hydrogen atom from the third energy level ($n_2 = 3$) to the second energy level ($n_1 = 2$). The energy of this photon $E_{p}$ (in eV) is given by the formula:
$E_{p} = 13.6 \left[\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right]$
Substituting $n_{1} = 2$ and $n_{2} = 3$:
$E_{p} = 13.6 \left[\frac{1}{2^{2}} - \frac{1}{3^{2}}\right] \\
\phantom{E_{p}}= 13.6 \left[\frac{1}{4} - \frac{1}{9}\right] \\
\phantom{E_{p}}= 13.6 \left[\frac{9 - 4}{36}\right] \\
\phantom{E_{p}}= 13.6 \times \frac{5}{36} \\
\phantom{E_{p}}= 1.89 \text{ eV}
$
Step 2: Relate the Work Function to Photon Energy and Kinetic Energy
When a photon with energy $E_{p}$ strikes the gold surface, some of this energy is used to overcome the work function ($\phi$) of the metal, and the rest appears as the kinetic energy ($KE_{\max}$) of the ejected electrons. Mathematically:
$\phi = E_{p} - KE_{\max} \quad \dots \text{(i)}$
Step 3: Calculate the Maximum Kinetic Energy from the Electron's Circular Motion
After being emitted, the electrons move in a magnetic field $B = 5 \times 10^{-4}\,\text{T}$ along a circular path of radius $r = 7 \times 10^{-3}\,\text{m}$. The charge on the electron is $q = 1.6 \times 10^{-19}\,\text{C}$ and its mass is $m = 9.1 \times 10^{-31}\,\text{kg}$. The velocity $v$ of each electron in a uniform magnetic field is given by:
$v = \frac{B \, q \, r}{m}
$
Substituting the given values:
$v = \frac{(5 \times 10^{-4}) \times (1.6 \times 10^{-19}) \times (7 \times 10^{-3})}{9.1 \times 10^{-31}} \\
\phantom{v}= 6.15 \times 10^{5}\,\text{m/s}
$
The kinetic energy $KE_{\max}$ in joules is:
$KE_{\max} = \frac{1}{2} m v^{2}
$
But we usually express $KE$ in eV. First, calculate in joules, then convert to eV by dividing by $1.6 \times 10^{-19}$. So:
$KE_{\max} = \frac{1}{2} \times 9.1 \times 10^{-31} \times \left(6.15 \times 10^{5}\right)^{2} \text{ J}
$
When converted to eV, one obtains approximately:
$KE_{\max} = 1.075 \,\text{eV}
$
Step 4: Calculate the Work Function of the Metal
Using equation (i),
$\phi = E_{p} - KE_{\max} \\
\phantom{\phi} = 1.89 \,\text{eV} - 1.075 \,\text{eV} \\
\phantom{\phi} = 0.82 \,\text{eV}
$
Hence, the work function $\phi$ of the gold surface is 0.82 eV.
