Question

If the average lifetime of an excited state of H atom is of order \[{10^{ - 8}}\] sec, then the no. Of orbits an \[{{\text{e}}^ - }\] makes when it is in the state \[{\text{n}} = 2\] before it suffers a transition to \[{\text{n}} = 2\] are:
A.\[8 \times {10^6}\]
B.\[9 \times {10^6}\]
C.\[7 \times {10^6}\]
D.\[6 \times {10^6}\]

Answer 1

Hint: First of all we will calculate the frequency of orbit, frequency if the reverse of the time period. The radius of the second state can be calculated using the Bohr formula and Bohr radius.

Formula used:
\[{{\text{r}}_{\text{n}}} = 0.529 \times {10^{ - 8}} \times {{\text{n}}^2}{\text{ cm}}\]
Here \[{{\text{r}}_{\text{n}}}\] is the radius of the nth orbit of hydrogen.
\[{{\text{V}}_{\text{n}}} = \dfrac{{2.185 \times {{10}^{ - 8}}}}{{\text{n}}}{\text{cm}}/{\text{s}}\]
Here \[{{\text{V}}_{\text{n}}}\] is the velocity of the nth orbit of hydrogen.

Complete step by step solution:
We know the basic relation between speed distance and time that is:
\[{\text{speed }} = \dfrac{{{\text{Distance}}}}{{{\text{Time}}}}\]
The reciprocal of time in called frequency. Hence we can re write the formula as:
\[{\text{speed }} = {\text{Distance}} \times {\text{frequency}}\]
Now since Bohr assumed the orbital to be circular in nature so the distance travelled by the electron in one rotation will be equal to the circumference of the circle which is \[2\pi {\text{r}}\] .
Hence we will put the value of speed and velocity:
\[\dfrac{{2.185 \times {{10}^{ - 8}}}}{{\text{n}}} = 2\pi \times 0.529 \times {10^{ - 8}} \times {{\text{n}}^2} \times {\text{frequency}}\]
The electron is revolving in the second orbit of hydrogen atom, hence the value of n will be 2. We will substitute the value as:
\[\dfrac{{2.185 \times {{10}^{ - 8}}}}{2} = 2\pi \times 0.529 \times {10^{ - 8}} \times 4 \times {\text{frequency}}\]
Solving the above equation we will get the frequency as:
\[8.2 \times {10^{14}}/{\text{ sec}}\]
That means in 1 sec the electron revolve \[8.2 \times {10^{14}}\] times in orbit. We need to calculate for \[{10^{ - 8}}\] sec. For \[{10^{ - 8}}\] sec the electron will revolve \[8.2 \times {10^{14}} \times {10^{ - 8}} = 8.2 \times {10^6}\]

Hence, the correct option is A.
Note:
Bohr’s model of an atom failed to explain the Zeeman Effect which is based on the magnetic field and stark effect which considers the electric field. It does not give account for spectra shown by species other than hydrogen and hydrogen like.