The problem involves the photoelectric effect, where ultraviolet (UV) light of energy is incident on a metal surface, causing electrons to be ejected. To find the maximum kinetic energy of these photoelectrons, we use the photoelectric equation given by Albert Einstein:
\(K_{\text{max}} = h \nu - \phi\)
Where:
Given:
Substituting these values into the equation:
\(K_{\text{max}} = 4.13 \text{ eV} - 3.13 \text{ eV}\)
\(K_{\text{max}} = 1 \text{ eV}\)
Thus, the maximum kinetic energy of the ejected photoelectrons is 1 eV.
Why other options are incorrect:
The correct answer is indeed 1 eV.
The energy of the ejected photoelectrons is given by the photoelectric equation:
\( K.E. = h\nu - \phi \),
where \( h\nu \) is the energy of the incident photons and \( \phi \) is the work function of the material.
Given:
\( h\nu = 4.13 \, \text{eV}, \quad \phi = 3.13 \, \text{eV} \),
the maximum kinetic energy of the ejected photoelectrons is:
\( K.E. = 4.13 \, \text{eV} - 3.13 \, \text{eV} = 1 \, \text{eV} \).
Thus, the correct answer is Option (2).
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.