Atoms And Nuclei Question 20

Question: An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given the Rydberg’s constant R=105cm1 . The frequency in Hz of the emitted radiation will be [CPMT 1976]

Options:

A) 316×105

B) 316×1015

C) 916×1015

D) 34×1015

Show Answer

Answer:

Correct Answer: C

Solution:

1λ=R,(122142)=3R16λ=163R=163×105cm

Frequency n=cλ=3×1010163×105=916×1015Hz



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