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Q.Number of electrons emitted from a piece of metal for giving 1×10−71 \times 10^{-7} coulomb charge will be:

i) 10710^{7}
ii) 1.6×10191.6 \times 10^{19}
iii) 6.25×10116.25 \times 10^{11}
iv) 9×10129 \times 10^{12}
Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2026MCQ· 1mImportance★★★★★
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Charge is quantised, so n=Q/e=6.25×1011n = Q/e = 6.25\times10^{11} electrons.

Concept. Charge exists only in whole multiples of the electronic charge e=1.6×10−19e = 1.6\times10^{-19} C (quantisation of charge). If a metal piece loses a charge QQ, the number of electrons removed is n=Qen = \dfrac{Q}{e}.

Calculation. …

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