All Free Physics MCQs with Answers

261. The momentum of a photon of wavelength lambda is

• A. h divided by lambda
• B. h times lambda
• C. zero, since a photon has no mass
• D. mc

Explanation: A photon has zero rest mass but still carries momentum h over lambda, which is what allows radiation pressure and the recoil seen in the Compton effect. Solar sails are propelled by exactly this momentum transfer. Assuming that no rest mass means no momentum is the standard misconception here.

262. The photoelectric effect is best demonstrated with which metal, given that it has a very low work function?

• A. Caesium
• B. Iron
• C. Tungsten
• D. Platinum

Explanation: Caesium's work function is only about 2 eV, so ordinary visible light is enough to eject electrons, whereas tungsten and platinum need ultraviolet. This is why alkali metals are used in photocells and photomultipliers. The low value reflects how loosely the single outer electron of an alkali metal is held.

263. A semiconductor doped with a pentavalent impurity such as phosphorus becomes

• A. n type, with electrons as the majority carriers
• B. p type, with holes as the majority carriers
• C. an insulator
• D. a conductor with no charge carriers

Explanation: A pentavalent atom has five valence electrons but only four are needed for bonding, so the fifth is loosely bound and becomes a free electron, making electrons the majority carriers. Doping with a trivalent impurity such as boron leaves a vacancy, a hole, and gives p type material. The crystal as a whole stays electrically neutral in both cases.

264. The majority charge carriers in a p type semiconductor are

• A. electrons
• B. holes
• C. protons
• D. neutrons

Explanation: Trivalent dopants create vacancies in the bonding structure, and these holes behave as mobile positive charges as neighbouring electrons shift to fill them. Electrons are still present as minority carriers, generated thermally. Protons and neutrons are locked in the nuclei and never move through the lattice.

265. The depletion region of a pn junction is a layer that

• A. contains a very high concentration of free carriers
• B. has been swept clear of mobile charge carriers, leaving fixed ions and a potential barrier
• C. conducts better than the rest of the crystal
• D. consists only of electrons

Explanation: Electrons diffusing across the junction recombine with holes, leaving exposed fixed ions on both sides and a region with almost no mobile carriers, so it acts as an insulator. The resulting internal field forms a potential barrier of about 0.7 V for silicon and 0.3 V for germanium. Forward biasing narrows this region and reverse biasing widens it.

266. A pn junction diode is forward biased when

• A. the p side is connected to the positive terminal and the n side to the negative
• B. the n side is connected to the positive terminal
• C. both sides are connected to the same terminal
• D. no voltage is applied

Explanation: Connecting positive to p repels holes and negative to n repels electrons, driving both towards the junction, so the depletion layer narrows and current flows once the applied voltage exceeds the barrier. Reverse bias does the opposite, widening the layer so that only a tiny leakage current passes. This one way behaviour is what makes the diode useful.

267. In reverse bias, the current through an ideal diode is

• A. large
• B. the same as in forward bias
• C. almost zero, apart from a small leakage current
• D. infinite

Explanation: The widened depletion region blocks the majority carriers, leaving only the thermally generated minority carriers to cross, which gives a leakage current of the order of microamperes. If the reverse voltage rises far enough the junction breaks down and conducts heavily, which is destructive for an ordinary diode but is the intended operating mode of a Zener diode. Ideal diode analysis simply treats the reverse current as zero.

268. The barrier potential across an unbiased silicon pn junction is about

• A. 0.3 V
• B. 0.7 V
• C. 5 V
• D. 12 V

Explanation: Silicon has a barrier of roughly 0.7 V, while germanium's is about 0.3 V, which is why a silicon diode needs about 0.7 V across it before it conducts appreciably. This forward voltage drop must be allowed for in circuit calculations. Silicon is preferred for most applications because it tolerates higher temperatures and has lower leakage.

269. Rectification is the process of converting

• A. direct current to alternating current
• B. alternating current to direct current
• C. low voltage to high voltage
• D. current to voltage

Explanation: A diode conducts in one direction only, so it can strip out one half of the alternating cycle and leave a unidirectional output. Converting direct to alternating current is inversion, done by an inverter, which is the opposite process. Every mains powered electronic device contains a rectifier of some kind.

270. In a half wave rectifier, the output

• A. uses both halves of the input cycle
• B. consists of one half cycle only, with the other half missing
• C. is a steady direct voltage
• D. is zero

Explanation: The single diode conducts only while its anode is positive, so alternate half cycles are blocked and the output is a series of pulses with gaps between them. This makes it inefficient and hard to smooth, since the ripple frequency is only that of the supply. A full wave rectifier inverts the other half instead of discarding it.