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Q.

If the electron falls from n = 3 to n = 2, in hydrogen atom then emitted energy is

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a

10.2 eV

b

12.09 eV

c

1.9 eV

d

0.65 eV

answer is C.

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Detailed Solution

E1=−13.6 E2=−3.4 E3=−1.51 E3−E2=−1.51+3.4=1.89 eV E3−E2=1.9 eV
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