For the endothermic reaction ⁷₃Li + ¹₁H → ⁷₄Be + ¹₀n (Q = −1645 keV), calculate the minimum incident proton kinetic energy. If it is 235n keV, find n. Lithium target is stationary.
Physics · Mixed Chapter
JEE Main Practice

Question

For the endothermic reaction ⁷₃Li + ¹₁H → ⁷₄Be + ¹₀n (Q = −1645 keV), calculate the minimum incident proton kinetic energy. If it is 235n keV, find n. Lithium target is stationary.

G1 Physics · Accepted Answer

Step 1 — Minimum kinetic energy for an endothermic two-body reaction on a stationary target is Eₜₕ = −Q (1 + mₚ/Mₜ). Step 2 — Li-7 target is approximately 7 proton masses, so Eₜₕ = 1645 × (1 + 1/7) keV. Step 3 — Eₜₕ = 1645 × 8/7 = 1880 keV. Step 4 — The source writes threshold = 235n keV; n = 1880/235 = 8.