Majority carrier in a $n −$ type semiconductor are holes. This statement is false.
Since, in $n−$type semiconductor, the pentavalent impurity atoms donate electrons o the host crystal and the semiconductor doped with donars $($pentavalent impurity$)$ is called $n −$ type semiconductor.
Therefore majority carrier in a $n−$type semiconductor are electrons.
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$A \to B + {\;_2}H{e^4}$
$B \to C + \;2{e^ - }$
Then