Solar Cell

8 MCQs9-step worked example
Source: NCERT Electronic DevicesPYQ coverage: NEET 2023, 2024Official key: NTA-verifiedLast updated: 26 Sep 2026

Solar Cell, explained for NEET

The confusion that costs marks here is treating a solar cell as a photodiode with a battery behind it. It is not. A solar cell is a p-n junction operated with no external bias at all — it is the source of emf, not a load. The photodiode's reverse-bias operation belongs to its own topic; hold the two apart.

NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 489 (NCERT removed the solar cell from the current book in 2023; the NEET (UG) 2026 syllabus still lists its I–V characteristics) gives the construction: a p-Si wafer about 300 μm thick, with a thin n-Si layer (~0.3 μm) diffused on one face. The p-side carries a metal back contact. The n-side carries a metal finger grid as front contact, covering under 15% of the top face so that light still reaches the junction. The junction area is deliberately large, because the aim is power, not mere detection.

Three processes generate the emf. Generation: photons with hν > E₉ create electron-hole pairs close to the junction. Separation: the built-in electric field of the depletion region sweeps electrons to the n-side and holes to the p-side. Collection: the front contact collects electrons, the back contact collects holes. The p-side becomes positive, the n-side negative — a photovoltage. Connect a load and a photocurrent flows through it.

The I-V characteristic is drawn in the fourth quadrant: the cell supplies current rather than drawing it, so the voltage is positive while the current is negative.

Material selection runs on five criteria: band gap roughly 1.0-1.8 eV, high optical absorption (~10⁴ cm⁻¹), adequate electrical conductivity, availability of raw material, and cost. Si (1.1 eV), GaAs (1.43 eV), CdTe (1.45 eV) and CuInSe₂ (1.04 eV) all fit. Solar radiation peaks near 1.5 eV, so a gap near or just below that harvests the most photons. GaAs edges out Si mainly on its higher absorption coefficient.

Watch-out: sunlight is not required. Any light with photon energy above E₉ works — the same cell powers a satellite and a calculator.

Can you answer these Solar Cell MCQs?

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

The I-V characteristic of an illuminated solar cell is conventionally drawn in which quadrant of the I-V plane?

Show answer and why every option is right or wrong

Answer: D. NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 489 places the solar cell characteristic in the fourth quadrant, because the cell supplies current to the load instead of drawing it — the voltage is positive while the current is negative.

Why A is wrong: A is wrong because the first quadrant (positive V, positive I) describes a junction that draws current from an external source; an illuminated solar cell delivers power rather than absorbing it.

Why B is wrong: B is wrong because the second quadrant needs negative voltage with positive current, which is not the polarity a photovoltaic cell develops — its p-side goes positive.

Why C is wrong: C is wrong because the third quadrant (negative V, negative I) describes a biased junction, and the solar cell in this topic carries no external bias at all.

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MCQ 2Easy RecallPractice

According to the NCERT selection criteria for solar cell materials, the band gap should lie approximately in which range?

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Answer: B. NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 489 lists a band gap of about 1.0-1.8 eV as the first criterion, which is why Si (1.1 eV), GaAs (1.43 eV), CdTe (1.45 eV) and CuInSe₂ (1.04 eV) are the materials named.

Why A is wrong: A is wrong because a gap that small lets almost every photon through with energy to spare, but the photovoltage a cell can develop is tied to the band gap, so very small gaps waste most of the absorbed energy.

Why C is wrong: C is wrong because a gap of 2.5-3.2 eV lies well above the ~1.5 eV peak of the solar spectrum, so the great majority of solar photons could not create electron-hole pairs at all.

Why D is wrong: D is wrong because 4.0-5.0 eV is insulator territory on the NCERT band-gap picture; such a solid would be transparent to essentially the whole solar spectrum.

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MCQ 3Easy RecallPractice

NCERT describes the generation of emf in a solar cell as three basic processes occurring in sequence. They are:

Show answer and why every option is right or wrong

Answer: C. NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 489 names the three processes as generation of electron-hole pairs by light, separation by the depletion-region field, and collection at the front and back contacts.

Why A is wrong: A is wrong because a solar cell neither amplifies nor rectifies in the photovoltaic mode described here; it converts optical power directly into electrical power with no external bias.

Why B is wrong: B is wrong because recombination is a loss mechanism that destroys the very carriers the cell needs; it is not one of the three emf-generating processes.

Why D is wrong: D is wrong because separation cannot precede generation — the electron-hole pairs must first be created by photons before the junction field can sweep them apart.

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MCQ 4Direct ApplicationPractice

In an illuminated solar cell, the photogenerated electrons and holes near the junction are driven in opposite directions mainly by:

Show answer and why every option is right or wrong

Answer: A. NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 489 states that the electrons and holes are separated by the electric field of the depletion region, electrons being swept to the n-side and holes to the p-side.

Why B is wrong: B is wrong because a solar cell is operated with no external bias whatsoever — that is precisely what distinguishes it from a light detector that is biased before use.

Why C is wrong: C is wrong because the finger grid is an ohmic front contact that collects carriers after they have already been separated; it plays no part in the separation itself.

Why D is wrong: D is wrong because unaided diffusion has no preferred direction and would not build a steady polarity across the cell; the junction field is what makes the separation directional.

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MCQ 5Direct ApplicationPractice

A solar cell is illuminated with its terminals left open. Which statement correctly describes the resulting terminal polarity and its cause?

Show answer and why every option is right or wrong

Answer: B. Following the separation and collection steps on NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 489, holes reaching the p-side are collected by the back contact and electrons reaching the n-side by the front contact, so the p-side becomes positive and the n-side negative.

Why A is wrong: A is wrong because it reverses the carrier destinations: the junction field sweeps electrons to the n-side, not the p-side, so the n-side ends up negative.

Why C is wrong: C is wrong because the photovoltage appears on open circuit — that is the defining feature of the photovoltaic effect. The load only allows a photocurrent to flow in addition.

Why D is wrong: D is wrong because no external bias is applied to a solar cell; the polarity is fixed by the direction of the junction's own built-in field.

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MCQ 6Direct ApplicationPractice

The front contact of a solar cell is made as a metal finger grid covering less than 15% of the top face, rather than as a continuous metal sheet. The reason is that:

Show answer and why every option is right or wrong

Answer: A. NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 489 states that the metal grid occupies only a very small fraction of the cell area so that light can be incident on the cell from the top, while still serving as the front contact.

Why B is wrong: B is wrong because the front contact touches only the n-Si layer; it does not bridge the p and n regions, so there is no short-circuit path across the junction whatever its area.

Why C is wrong: C is wrong because band gap is a property of the semiconductor material, unaffected by the shape of a metal electrode deposited on its surface.

Why D is wrong: D is wrong because the ohmic quality of a metal-semiconductor contact depends on the materials and doping, not on whether the metal is laid down as strips or as a sheet.

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MCQ 7CalculationPractice

Two solar cells of identical junction area are exposed to the same sunlight. Cell X is made from a semiconductor of band gap 1.1 eV and cell Y from one of band gap 2.4 eV. Which cell generates the larger photocurrent, and why?

Show answer and why every option is right or wrong

Answer: D. Electron-hole pairs are created only by photons with hν greater than the band gap (NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 490), and since the solar spectrum peaks near 1.5 eV, many more photons clear the 1.1 eV threshold than the 2.4 eV one, giving cell X the larger photocurrent.

Why A is wrong: A is wrong because it answers the wrong question: a larger gap can raise the voltage per carrier, but the photocurrent is set by how many photons exceed the threshold, and at 2.4 eV most of the solar spectrum is below it.

Why B is wrong: B is wrong because equal incident power does not mean equal photocurrent — photons below the band gap pass through without creating any carriers, so part of that power is simply not converted.

Why C is wrong: C is wrong because reduced thermalisation loss is a statement about energy per carrier, not about carrier number; it cannot make up for the photons the 2.4 eV material fails to absorb.

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MCQ 8Concept TrapPractice

A student says: "A solar cell is just a light detector reverse-biased by a battery, given a bigger window." Which single statement corrects this claim?

Show answer and why every option is right or wrong

Answer: C. NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 489 defines a solar cell as a p-n junction that generates emf when solar radiation falls on it, with no external bias applied and with the junction area kept much larger because more power is wanted.

Why A is wrong: A is wrong on both counts: no bias of any kind is applied, and the junction area is deliberately enlarged, which is the second half of the definition.

Why B is wrong: B is wrong because NCERT states the opposite — sunlight is not always needed, and any light with photon energy above the band gap will generate the emf.

Why D is wrong: D is wrong because a forward bias would make the cell draw power from a source; the illuminated cell delivers power, which is why its characteristic lies in the fourth quadrant.

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How do you solve a Solar Cell question? A worked example

  1. 1

    Given

    • Top face area of one solar cell, A = 1.00 × 10² cm²• Fraction of the top face shaded by the metal finger grid, f = 12% (exact design figure)• Incident solar power density, S = 1.0 × 10³ W m⁻² (normal incidence)

  2. 2

    Required

    The optical power actually reaching the active n-Si surface of the cell.

  3. 3

    Concept

    NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 489 notes that the front contact is a metal grid occupying only a very small fraction of the cell area, so that light can be incident on the cell from the top. The metal covers that fraction opaquely; only the uncovered fraction of the top face admits light to the junction.

  4. 4

    Formula

    P = S × A_active, where A_active = (1 − f) A.

  5. 5

    Substitution

    A = 1.00 × 10² cm² × 10⁻⁴ m² cm⁻² = 1.00 × 10⁻² m²
    A_active = (1 − 0.12) × 1.00 × 10⁻² m² = 0.88 × 10⁻² m²
    P = (1.0 × 10³ W m⁻²) × (0.88 × 10⁻² m²)

  6. 6

    Calculation

    P = 1.0 × 10³ × 0.88 × 10⁻² = 8.8 W

    The conversion factor 10⁻⁴ m² per cm² is an exact definition, and the shading fraction is given as an exact design percentage, so neither of them contributes to the significant-figure count. Only the measured quantities do.

  7. 7

    Final answer

    P = 8.8 W, quoted to two significant figures because the incident power density 1.0 × 10³ W m⁻² is the least precise measured input.

  8. 8

    Common trap

    Using the whole 1.00 × 10² cm² face and reporting 10 W — the grid shadow is ignored. The second habitual slip is forgetting the cm² → m² conversion, which inflates the answer by a factor of 10⁴.

  9. 9

    Similar NEET-style question

    A solar cell of top-face area 25 cm² has a front finger grid shading 8.0% of that face. Sunlight of power density 1.0 × 10³ W m⁻² falls normally on it. Find the optical power reaching the active surface. (Answer: 2.3 W.)

What to remember before solving Solar Cell questions

Key Fact

Solar cell

In the NEET syllabus; removed from current NCERT.

A p-n junction that generates emf when solar radiation falls on it, working on the photovoltaic effect like a photodiode but with no external bias and a much larger junction area. Construction: a p-Si wafer of about 300 µm with a thin (~0.3 µm) n-Si layer diffused on one side; metal back contact on the p-side; a metallic grid front contact on the n-side occupying under 15% of the area. The emf arises from generation of e-h pairs by light (hν > E_g) near the junction, their separation by the depletion-region field, and their collection (electrons at the front contact, holes at the back contact). Its I-V characteristic is drawn in the fourth quadrant, because it supplies current to the load. Material criteria: band gap ~1.0 to 1.8 eV, high optical absorption, electrical conductivity, availability and cost (Si 1.1 eV, GaAs 1.43 eV, CdTe 1.45 eV, CuInSe₂ 1.04 eV).

-- NCERT Class 12 Physics (pre-2023 edition), Chapter 14, p. 489
Key Fact

Optoelectronic junction devices

In the NEET syllabus; removed from current NCERT.

Semiconductor diodes in which carriers are generated by photons (photo-excitation): (i) photodiodes, used for detecting optical signals (photodetectors); (ii) light emitting diodes (LED), which convert electrical energy into light; (iii) photovoltaic devices, which convert optical radiation into electricity (solar cells).

-- NCERT Class 12 Physics (pre-2023 edition), Chapter 14, p. 486

More in Electronic Devices: 3 exam traps and mistakes · 3 formulas · 3 question patterns from its other lessons.

Solar Cell questions from past NEET papers

2 questions from NEET 2023, 2024. Answers verified against NTA official keys. — click to collapse
NEET 2023

Given below are two statements: Statement I: Photovoltaic devices can convert optical radiation into electricity. Statement II: Zener diode is designed to operate under reverse bias in breakdown region. In the light of the above statements, choose the most appropriate answer from the options given below.

1Both Statement I and Statement II are incorrect
2Statement I is correct but Statement II is incorrect
3Statement I is incorrect but Statement II is correct
4Both Statement I and Statement II are correct
NTA Answer: Option 4(final)

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All 17 past-paper questions from Electronic Devices →

Sources

Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.

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SemiconductorsZener Diode