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)
- A \( \frac{3 R C}{27} \)
- B \( \frac{R C}{25} \)
- C \( \frac{8 R C}{9} \)
- D \( \frac{5 R C}{36} \)
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 \( \frac{5}{36} \mathrm{Rc} \)
\(\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \)
\( \text {Using } \lambda f=c \Rightarrow \frac{1}{\lambda}=\frac{f}{c} \)
\( \frac{f}{c}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \Rightarrow f=R c\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \)
\( \text {Given } n_{1}=2 ; n_{2}=3 \)
\( \Rightarrow f=R c\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R c\left(\frac{1}{4}-\frac{1}{9}\right)=R c \frac{(9-4)}{4 \times 9}=\frac{5 R C}{36}\)
Therefore, frequency of emitted radiation is \( \frac{5}{36} \mathrm{Rc} \)
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