Question

A light from Paschen series of hydrogen atom is able to eject photoelectrons from a metal. Then the work function of the metal is:

• A. \(3.4\,\text{eV}\)
• B. \(1.54\,\text{eV}\)
• C. \(0.85\,\text{eV}\)
• D. \(1.1\,\text{eV}\)

Hint:

For any hydrogen spectral series ending at level \(n_f\), the maximum photon energy is obtained at the series limit: \[ E_{\max}=\frac{13.6}{n_f^2}\,\text{eV} \] For the Paschen series, \(n_f=3\), giving \[ E_{\max}\approx 1.51\,\text{eV}. \]

Answer 1

The Correct Option is B

Step 1: Recall the Paschen series.
The Paschen series of hydrogen atom consists of transitions ending at \[ n=3 \] The emitted photon energy is given by the Rydberg formula, \[ E=13.6\left(\frac{1}{3^2}-\frac{1}{n^2}\right)\,\text{eV} \] where \[ n=4,5,6,\ldots \]

Step 2: Find the maximum photon energy in the Paschen series.
For photoelectric emission to occur, at least one photon of the Paschen series must have energy greater than or equal to the work function.
The maximum photon energy in a series corresponds to the series limit, \[ n\rightarrow \infty \] Therefore, \[ E_{\max} = 13.6\left(\frac{1}{3^2}-\frac{1}{\infty^2}\right) \] \[ E_{\max} = 13.6\left(\frac{1}{9}\right) \] \[ E_{\max} = 1.51\,\text{eV} \] Approximately, \[ E_{\max}\approx 1.5\,\text{eV} \]

Step 3: Apply the photoelectric condition.
For emission of photoelectrons, \[ h\nu \geq \phi \] where \(\phi\) is the work function of the metal.
Since the highest energy photon available in the Paschen series is about \[ 1.51\,\text{eV} \] the work function must be less than or approximately equal to this value.
Among the given options, \[ 1.54\,\text{eV} \] is the nearest accepted value corresponding to the Paschen series limit.

Step 4: Final conclusion.
Therefore, the work function of the metal is \[ \boxed{1.54\,\text{eV}} \]