© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Photon Energy from the 3 → 2 Transition
The energy of the photon emitted when the electron in a hydrogen atom undergoes a transition from the third to the second energy level is given by:
$E_{\text{photon}} = 13.6\,eV \left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right)$
Evaluate this difference:
$\dfrac{1}{4} - \dfrac{1}{9} = \dfrac{9 - 4}{36} = \dfrac{5}{36}
\quad\Longrightarrow\quad
E_{\text{photon}} = 13.6 \times \dfrac{5}{36} = 1.88\,eV \, (\text{approximately}).$
Step 2: Relate the Electron's Kinetic Energy to the Radius of Circular Path in a Magnetic Field
After absorbing the photon, electrons are ejected from the metal with some maximum kinetic energy $K_{\max}$. These electrons move in a magnetic field $B = 3 \times 10^{-4}\,T$ and follow a circular path of radius $r = 10.0\,\text{mm} = 10.0 \times 10^{-3}\,\text{m}.$ The radius $r$ of an electron’s circular path in a uniform magnetic field is given by:
$r = \dfrac{m v}{q B},
$
where $m$ is the electron mass, $v$ is its speed, and $q$ is the electron charge ($q = e$). Because $K_{\max} = \frac{1}{2} m v^2 = eV$ (if $V$ is the potential equivalent of the kinetic energy in electron-volts), one can rewrite the relation to find:
$r = \dfrac{\sqrt{2 m e V}}{e B} \;\;\Longrightarrow\;\;
V = \dfrac{B^2\, r^2\, e}{2\,m}.
Step 3: Calculate the Electron Kinetic Energy
Plug the given values ($B,$ $r,$ $m,$ $e$) into the expression to find the electron’s kinetic energy (in eV). Performing that calculation gives approximately:
$K_{\max} = eV \approx 0.8\,eV.
$
Step 4: Determine the Metal’s Work Function
The maximum kinetic energy of the photoelectrons is the difference between the photon energy and the metal’s work function $\phi$. Hence,
$\phi = E_{\text{photon}} - K_{\max}.
$
Substitute the numerical values:
$\phi \approx 1.88\,eV - 0.8\,eV = 1.08\,eV.
$
Rounding suitably, we get:
$\phi \approx 1.1\,eV.
$
Step 5: Conclusion
Thus, the work function of the metal is closest to $1.1\,eV.$
