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एक तत्त्व के ऋणायन `X^(-)` में 18 इलेक्ट्रान हैं । तत्व की द्रव्यमान संख्या 35 है । तत्व के नाभिक में न्यूट्रॉनों की सांख्या निकालो ।

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ऋणायन `(X^(-))` में एलेक्ट्रोनो की संख्या = 18
अतः तत्व की परमाणु ( X ) में एलेक्ट्रॉनों की संख्या = 18 - 1 = 17
तत्व के नाभिक में प्रोटॉन की संख्या उसके परमाणु में उपस्थित एलेक्ट्रॉनों की संख्या ( अर्थात 17 ) के बराबर होगी ।
A = Z + n
A= 35, Z = 17
`:. ` n = A-Z = 35-17 = 18 ( न्यूट्रॉन की संख्या )

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