Two parallel metallic plates each have area 5 × 10⁻⁴ m² and are 1 cm apart. Plate B has charge +33.7 × 10⁻¹² C. Photons of energy 5 eV are incident on plate A at 10¹⁶ photons m⁻² s⁻¹; one electron is emitted per 10⁶ incident photons. The work function is 2 eV. Find x if number of emitted electrons before t = 10 s is x × 10⁷. [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

Two parallel metallic plates each have area 5 × 10⁻⁴ m² and are 1 cm apart. Plate B has charge +33.7 × 10⁻¹² C. Photons of energy 5 eV are incident on plate A at 10¹⁶ photons m⁻² s⁻¹; one electron is emitted per 10⁶ incident photons. The work function is 2 eV. Find x if number of emitted electrons before t = 10 s is x × 10⁷. [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Number of incident photons per second is flux × illuminated area: 10¹⁶ × 5 × 10⁻⁴ = 5 × 10¹² photons s⁻¹. Step 2 — Photoemission yield is one electron per 10⁶ photons: electron rate 5 × 10⁶ s⁻¹. Step 3 — During 10 s, N = 5 × 10⁷ electrons. Step 4 — In N = x × 10⁷, x = 5. [The charge/voltage data affect collection, not the specified emission count under the stated idealization.]