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An ion with mass number 79 when placed between two electrically charged plates with potential difference 1 V gains energy of 2 eV and moves towards the positive plate. If the ion contains 25 % more neutrons than the electrons, identify the ion.

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Energy = 2eV, Voltage = 1V

y = 36

then number of proton = y – 2

= 36 – 2 = 34

So, atomic number of X will be = 34 & Ion will be = Se–2

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