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QAIPM-2012-Ph-1764: NEET 2012 Physics - Class 12 AIPM-2012-Ph-1764 Solution: An electron of a stationary hydrogen atom passes from the fifth …

An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be

\frac{24hR}{25m}

\frac{25hR}{24m}

\frac{24m}{25hR}

\frac{25m}{24hR}

Multiple Choice Options

A 4 $\times$ _ 10 15 Hz
B 5 $\times$ _ 10 15 Hz
C 1.6 $\times$ _ 10 15 Hz
D 2.5 $\times$ _ 10 15 Hz

Step-by-Step Verified Solution

Energy released from first exited state to the ground state : E = (� 3.4) � (� 13.6) = 10.2 eV Work function _ = E � 3.75 eV = (10.2 � 3.75) eV = 6.63 eV _ h _ 0 = 6.63 _ _ 0 = 6.63 _ 1.6 _ 10 _ 19 6.63 _ 10 _ 34 = 1.6 _ 10 15 Hz

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