A one‑dimensional box of length $L = 0.3 nm$ contains an electron. Using the particle‑in‑a‑box model, calculate the ground state energy $E_{1}$ in electron‑volts. Use $h = 6.626 \times 1 0^{- 34} J\cdots$ and $m_{e} = 9.11 \times 1 0^{- 31} kg$ .

medium 2 marks Class 12 · Physics · Atoms

Answer

4.2 eV

The correct answer is 4.2 eV — Step 6: Convert to eV by dividing by $1.602 \times 1 0^{- 19}$ : $E_{1} = 6.69 \times 1 0^{- 19} /1.602 \times 1 0^{- 19} \approx 4.18 eV$ , round to $4.2 eV$ .

Step-by-step solution

Step 1: Recall the ground state energy formula $E_{1} = \frac{h ^{2}}{8 m L ^{2}}$ for a 1‑D box.
Step 2: Convert the length to SI units: $L = 0.3 nm = 0.3 \times 1 0^{- 9} m$ .
Step 3: Compute $h^{2} = (6.626 \times 1 0^{- 34})^{2} = 4.39 \times 1 0^{- 67} J^{2} s^{2}$ .
Step 4: Compute the denominator $8 m L^{2} = 8 \times 9.11 \times 1 0^{- 31} \times (0.3 \times 1 0^{- 9})^{2} = 6.56 \times 1 0^{- 49} kg\cdotm^{2}$ .
Step 5: Divide to get $E_{1} = 4.39 \times 1 0^{- 67} /6.56 \times 1 0^{- 49} = 6.69 \times 1 0^{- 19} J$ .
Step 6: Convert to eV by dividing by $1.602 \times 1 0^{- 19}$ : $E_{1} = 6.69 \times 1 0^{- 19} /1.602 \times 1 0^{- 19} \approx 4.18 eV$ , round to $4.2 eV$ .

Common mistake: Using

E_{n} = \frac{n ^{2} h ^{2}}{8 m L ^{2}}

for

n = 1

incorrectly or forgetting the square of

L

Concept tested: Particle‑in‑a‑box ground state energy

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

← All Atoms questions

A one‑dimensional box of length L=0.3 nm contains an electron. Using the particle‑in‑a‑box model, calculate the ground state energy E1​ in electron‑volts. Use h=6.626×10−34 J⋅s and me​=9.11×10−31 kg.

Step-by-step solution

A one‑dimensional box of length $L = 0.3 nm$ contains an electron. Using the particle‑in‑a‑box model, calculate the ground state energy $E_{1}$ in electron‑volts. Use $h = 6.626 \times 1 0^{- 34} J\cdots$ and $m_{e} = 9.11 \times 1 0^{- 31} kg$ .