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
A
316×105
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B
316×1015
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C
916×1015
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D
34×1015
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Solution
The correct option is C916×1015 1λ=R(122−142)=3R16⇒λ=163R=163×10−5cm
Frequency n = cλ=3×1010163×10−5=916×1015Hz