1.

What is the symbol for the nucleus remaining after ._(20)Ca^(42) undergoes beta-emission

Answer»

`._(21)Ca^(42)`
`._(20)SC^(42)`
`._(21)Sc^(42)`
`._(21)Sc^(41)`

Solution :`._(20)Ca^(42) rarr ._(21)Sc^(42) + ._(-1)E^(0)`


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