Given n=2,Z=1 2πr=nλ2π×(0.052Zn2)=nλ On solving λ=0.67 nm
De-Broglie wavelength of an electron orbiting in the n=2 state of hydrogen atom is close to (Given Bohr radius =0.052 nm )
Held on 30 Apr 2025 · Verified 9 Jul 2026.
0.067 nm
0.67 nm
1.67 nm
2.67 nm
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