question_answer

(a) Using Bohr's second postulate of quantization of orbital angular momentum, show that the circumference of the electron in the \[nth\] orbital state in hydrogen atom is \[n\] times the de-Broglie wavelength associated with it. (b) The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?                                  

Answer:

                (a) According to Bohr's quantisation condition, \[L=m\upsilon {{r}_{n}}=\frac{nh}{2\pi },n=1,2,3...\]                 or                            \[2\pi {{r}_{n}}=n\frac{h}{m\upsilon }\]                 But         \[\frac{h}{m\upsilon }=\]de- Broglie wavelength \[(\lambda )\]                 \[\therefore \]  \[2\pi {{r}_{n}}=n\lambda \] Thus the circumference of \[nth\]orbit contains exactly n de- Broglie wavelengths. (b) For third excited state, \[n=4\] For ground state, \[n=1\] Hence, the possible transitions are \[{{n}_{i}}=4\] to \[{{n}_{f}}=3,2,1\] \[{{n}_{i}}=3\]to \[{{n}_{f}}=2,1\] \[{{n}_{i}}=2\] to \[{{n}_{f}}=1\] \[\therefore \]Total number of transitions \[=6,\]as shown in Fig.