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An electron jumps from the 4th orbit to 2nd orbit of hydrogen atom. Given the Rydberg's constant the frequency in Hz of there emitted radiation will be
detailed solution
Correct option is C
1λ=R1n12−1n22 v=cλ=CR1n12−1n22 =3×1081071(2)2−1(4)2 =3×1081071(2)2−1(4)2 =916×1015HzTalk to our academic expert!
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