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Correct option is B)

$E_{Ph}=K.E_{max}+W$

$=eV_{0}+W=10+2.75=12.75eV$

Differenced of 4 and 1 energy level is 12.75 eV

So higher energy level is 4 to ground and Excited state is $n=3.$

$=eV_{0}+W=10+2.75=12.75eV$

Differenced of 4 and 1 energy level is 12.75 eV

So higher energy level is 4 to ground and Excited state is $n=3.$

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