NEET 2025 — Physics solution

NEET2025Physics

De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to (Given Bohr radius $= 0.052$ nm):

Solution

For the Bohr orbit, the circumference equals $n$ de-Broglie wavelengths: $2 π r = nλ$ .

For hydrogen ( $Z = 1$ ), $r = 0.052 \cdot n^{2} / Z = 0.052 \times 4 = 0.208$ nm.

$λ = \frac{2 π r}{n} = \frac{2 π \times 0.208}{2} \approx 0.67$ nm.

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