Question
Download Solution PDFहाइड्रोजन परमाणु की n = 2 अवस्था में परिक्रमा करने वाले इलेक्ट्रॉन की दे-ब्रॉग्ली तरंगदैर्ध्य क्या है?
(दिया गया है: बोर त्रिज्या = 0.052 nm)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसही विकल्प: (2) 0.67 nm है।
दिया गया है n = 2, Z = 1
2πr = nλ
2π × (0.052 × n2 / Z) = nλ
हल करने पर, λ = 0.67 nm
Last updated on Jun 16, 2025
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