Question

In the following transitions, which one has higher frequency?

• A. $3\to 1$
• B. $4\to 2$
• C. $4\to 3$
• D. $3\to 2$

Answer 1

Answer: (A)

$3\to 1$

Answer 2

From Bohr's postulate, energy of electron in $n t h$ orbit is given by $E=-\frac{M Z^{2} e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\left(\frac{1}{n^{2}}\right)$ When electron jumps from some higher energy state $n_{2}$ to a lower energy state $n_{1}$, the energy difference between these states is $E_{2}-E_{1} \propto\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$ From Bohr's third postulate, the frequency $v$ of the electromagnetic wave is $v=\frac{E_{2}-E_{1}}{h} \propto\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$ First case $n_{1}=1, n_{2}=3$ $\therefore v_{1} \propto\left(1-\frac{1}{9}\right) \propto \frac{8}{9}$ Second case $n_{1}=2, n_{2}=4$ $\therefore v_{2} \propto\left(\frac{1}{4}-\frac{1}{16}\right) \propto \frac{3}{16}$ Third case $n_{1}=3, n_{2}=4$ $\therefore v_{3} \propto\left(\frac{1}{9}-\frac{1}{16}\right) \propto \frac{7}{144}$ Fourth case $n_{1}=2, n_{2}=3$ $\therefore v_{4} \propto\left(\frac{1}{4}-\frac{1}{9}\right) \propto \frac{5}{36}$ $v_{1}>v_{2}>v_{4}>v_{3}$ Hence, transition $3 \rightarrow 1$ has higher frequency.