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KCET · Physics · Alternating Current

If an electron in hydrogen atom jumps from an orbit of level \( n=3 \) to an orbit of level \( n=2 \). the emitted radiation has a frequency ( \( R= \) Rydberg constant, \( C= \) velocity of light)

  1. A \( \frac{3 R C}{27} \)
  2. B \( \frac{R C}{25} \)
  3. C \( \frac{8 R C}{9} \)
  4. D \( \frac{5 R C}{36} \)
Verified Solution

Answer & Solution

Correct Answer

(D) \( \frac{5 R C}{36} \)

Step-by-step Solution

Detailed explanation

We know that
\(\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \)
\( \text {Using } \lambda f=c \Rightarrow \frac{1}{\lambda}=\frac{f}{c} \)
\( \frac{f}{c}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \Rightarrow f=R c\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \)
\( \text {Given } n_{1}=2 ; n_{2}=3 \)
\( \Rightarrow f=R c\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R c\left(\frac{1}{4}-\frac{1}{9}\right)=R c \frac{(9-4)}{4 \times 9}=\frac{5 R C}{36}\)
Therefore, frequency of emitted radiation is \( \frac{5}{36} \mathrm{Rc} \)