An electron jumps from the 4 th orbit to the 2 nd orbit of hydrogen atom. Given the…
Physics · JEE Advanced · NTA Exams — Atoms and Nuclei
An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given the Rydberg’s constant R = 105 cm–1. The frequency in Hz of the emitted radiation will be
- \(\frac{3}{6} \times 0^{5}\)
- \(\frac{3}{6} \times 0^{5}\)
- \(\frac{9}{6} \times 0^{5}\)
- \(\frac{3}{4} \times 0^{5}\)
Answer
(C) 9 6 0^ 5
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