An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given the Rydberg's constant R=105cm-1. The frequency in Hz of the emitted radiation will be
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a
316×105
b
316×1015
c
916×1015
d
34×1015
answer is C.
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Detailed Solution
1λ=R 122-142=3R16⇒λ=163R=163×10-5cm Frequency n=cλ=3×1010163×10-5=916×1015Hz