Option 1: Full wave rectification using a bridge of four diodes
Step 1 (Idea): Rectification means letting current pass in only one direction. A diode does exactly this because it conducts under forward bias and blocks under reverse bias. A full wave rectifier arranges diodes so that current reaches the load the same way during every half of the AC cycle.
Step 2 (Bridge circuit): Connect four diodes \(D_1, D_2, D_3, D_4\) in a diamond (bridge). The AC secondary of the transformer feeds the two opposite corners of the diamond, and the load \(R_L\) is connected across the other two corners. No centre tap is needed.
Step 3 (First half cycle): When the upper input terminal is positive, diodes \(D_1\) and \(D_3\) become forward biased and conduct, while \(D_2\) and \(D_4\) stay off. Current runs from the source through \(D_1\), down through \(R_L\), and returns through \(D_3\).
Step 4 (Second half cycle): When the polarity reverses, \(D_2\) and \(D_4\) conduct while \(D_1\) and \(D_3\) are off. Current now flows through \(D_2\), again downward through \(R_L\) in the same direction as before, and returns through \(D_4\).
Step 5 (Outcome): Because the load current keeps the same direction in both half cycles, the AC input is fully rectified into pulsating DC. Adding a capacitor filter smooths the ripples.
\[\boxed{\text{Bridge of 4 diodes} \Rightarrow \text{both half cycles rectified.}}\]
Option 2: Photoelectric energy via the \(hc = 1240\) eV nm shortcut
Step 1 (Matter waves): de Broglie's hypothesis says a moving particle behaves like a wave of wavelength \(\lambda = h/p\), where \(p\) is its momentum. So particles like electrons show diffraction, just as waves do; the heavier and faster the particle, the shorter its matter wavelength.
Step 2 (Photon energy shortcut): A handy form of \(E = hc/\lambda\) is
\[ E(\text{eV}) = \frac{1240}{\lambda(\text{nm})}. \]
Here \(\lambda = 3500\) Angstrom \(= 350\) nm, so
\[ E = \frac{1240}{350} \approx 3.54\ \text{eV}. \]
Step 3 (Einstein's equation): The maximum kinetic energy of a photoelectron is the photon energy left over after paying the work function \(W_0\):
\[ KE_{\max} = E - W_0. \]
Step 4 (Substitute): With \(W_0 = 2.0\) eV,
\[ KE_{\max} = 3.54 - 2.0 = 1.54\ \text{eV}. \]
Step 5 (Check): The photon energy exceeds the work function, so emission does occur, and the surplus \(1.54\) eV appears as the fastest electrons' kinetic energy.
\[\boxed{KE_{\max} \approx 1.54\ \text{eV}}\]