Question

How is the p-n junction diode used as a full wave rectifier? Explain its working process by making its simple circuit.
OR
What are de Broglie's matter-waves? The work function of potassium is \( 2\!\cdot\!0\ \text{eV} \). When ultraviolet light of wavelength \( 3500\ \text{\AA} \) falls on the surface of potassium, calculate the maximum kinetic energy of the emitted photoelectrons in electron-volt.

Hint:

For the rectifier, use that a diode conducts only in forward bias; two diodes with a centre-tapped transformer (or a 4-diode bridge) send current through the load the same way in both half cycles. For the OR part, use \( KE_{max} = \frac{hc}{\lambda} - W_0 \).

Answer 1

Option 1: p-n junction diode as a full wave rectifier

Step 1 (Principle): A p-n junction diode conducts current only when it is forward biased (p-side positive) and blocks current when reverse biased. A rectifier uses this one-way property to convert alternating current (AC) into direct current (DC). A full wave rectifier delivers output current during both halves of the AC input.
Step 2 (Circuit): Use a transformer whose secondary is centre-tapped at point \( O \). The two ends \( A \) and \( B \) of the secondary are connected to the p-sides of two diodes \( D_1 \) and \( D_2 \). The n-sides of both diodes are joined together and connected through the load resistance \( R_L \) back to the centre tap \( O \). The output is taken across \( R_L \). (Alternatively a bridge of four diodes may be used.)
Step 3 (Positive half cycle): During the half cycle when end \( A \) is positive and \( B \) is negative, diode \( D_1 \) is forward biased and conducts, while \( D_2 \) is reverse biased and does not conduct. Current flows from \( A \) through \( D_1 \), then through \( R_L \), and back to the centre tap \( O \).
Step 4 (Negative half cycle): During the next half cycle \( B \) becomes positive and \( A \) negative; now \( D_2 \) is forward biased and conducts while \( D_1 \) is off. Current flows from \( B \) through \( D_2 \), then through \( R_L \), and back to \( O \).
Step 5 (Result): In both half cycles the current through the load \( R_L \) flows in the same direction. Therefore both halves of the AC input are rectified, and the output is a pulsating direct current (a smoother DC than a half-wave rectifier gives). A filter capacitor across \( R_L \) further smooths this output.
\[\boxed{\text{Both half cycles drive current through } R_L \text{ in one direction} \Rightarrow \text{full wave rectification.}}\]

Option 2: de Broglie matter-waves and photoelectric energy

Step 1 (Matter waves): de Broglie proposed that every moving particle has a wave associated with it, called a matter wave. A particle of momentum \( p = mv \) has a wavelength
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \] where \( h \) is Planck's constant. This wave nature is significant only for very light particles such as electrons.
Step 2 (Photoelectric formula): By Einstein's photoelectric equation, the maximum kinetic energy of an emitted photoelectron is
\[ KE_{\max} = E_{\text{photon}} - W_0 = \frac{hc}{\lambda} - W_0 \] where \( W_0 \) is the work function.
Step 3 (Energy of the photon): With \( h = 6\!\cdot\!6\times10^{-34}\ \text{J s} \), \( c = 3\times10^{8}\ \text{m/s} \) and \( \lambda = 3500\ \text{\AA} = 3\!\cdot\!5\times10^{-7}\ \text{m} \),
\[ E = \frac{hc}{\lambda} = \frac{(6\!\cdot\!6\times10^{-34})(3\times10^{8})}{3\!\cdot\!5\times10^{-7}} = \frac{1\!\cdot\!98\times10^{-25}}{3\!\cdot\!5\times10^{-7}} = 5\!\cdot\!66\times10^{-19}\ \text{J} \]
Step 4 (Convert to eV): Dividing by \( 1\!\cdot\!6\times10^{-19}\ \text{J/eV} \),
\[ E = \frac{5\!\cdot\!66\times10^{-19}}{1\!\cdot\!6\times10^{-19}} \approx 3\!\cdot\!54\ \text{eV} \]
Step 5 (Maximum kinetic energy):
\[ KE_{\max} = 3\!\cdot\!54 - 2\!\cdot\!0 = 1\!\cdot\!54\ \text{eV} \]
\[\boxed{KE_{\max} \approx 1\!\cdot\!54\ \text{eV}}\]