In a hydrogen atom, an electron makes transition from $ n=3 $ to $ n=1 $ state in time interval of $ 1.2\times {{10}^{-8}}s $ , calculate average torque $ \left( Nm \right) $ acting on the electron during this transition.
$ \begin{align}
& \left( A \right)1.055\times {{10}^{-26}} \\
& \left( B \right)4.40\times {{10}^{-27}} \\
& \left( C \right)1.7\times {{10}^{-26}} \\
& \left( D \right)8.79\times {{10}^{-27}} \\
\end{align} $
Answer
562.2k+ views
Hint :In order to solve this question, we are going to first calculate the angular momentum of the electron in both the shells $ 2 $ and $ 3 $ by referring to the Bohr’s model of atom, and then by finding the differences of the angular momenta of the two shells and then dividing by time, we get the torque.
The formulae that are used here are:
According to the Bohr’s atomic model, the angular momentum of an electron present in a shell n is given by
$ L=\dfrac{nh}{2\pi } $
And the torque of an electron is given by
$ \begin{align}
& \tau =\dfrac{dL}{dt} \\
& \Rightarrow \tau =\dfrac{{{L}_{f}}-{{L}_{i}}}{{{t}_{f}}-{{t}_{i}}} \\
\end{align} $ .
Complete Step By Step Answer:
First of all, we must see that the time interval for the transition is already given i.e. $ 1.2\times {{10}^{-8}}s $ , now according to Bohr’s model of atom, the angular momentum for the various shells is given by
$ L=\dfrac{nh}{2\pi } $
For $ n=2 $ ,
$ {{L}_{2}}=\dfrac{2h}{2\pi }=\dfrac{h}{\pi } $
And for $ n=3 $
$ {{L}_{3}}=\dfrac{3h}{2\pi } $
Now as we know that the torque can be found from the angular momentum by the relation,
$ \tau =\dfrac{dL}{dt} $
i.e., for the case of transition of electron
$ \begin{align}
& \tau =\dfrac{{{L}_{2}}-{{L}_{3}}}{t} \\
& \Rightarrow \tau =\dfrac{\dfrac{3h}{2\pi }-\dfrac{2h}{2\pi }}{1.2\times {{10}^{-8}}} \\
& \Rightarrow \tau =\dfrac{\dfrac{h}{2\pi }}{1.2\times {{10}^{-8}}} \\
& \Rightarrow \tau =\dfrac{6.626\times {{10}^{-34}}}{2\times 3.14\times 1.2\times {{10}^{-8}}}=0.8792\times {{10}^{-26}} \\
& \Rightarrow \tau =8.79\times {{10}^{-27}} \\
\end{align} $
Hence, option $ \left( D \right)8.79\times {{10}^{-27}} $ is correct.
Note :
It is very important to see that the change in time $ dt $ given in the formula for the torque and angular momentum relation, is the time interval only, which is given in the question. The angular momentum of an electron in the atom depends directly on the shell in which the electron is present and the torque depends on the transition.
The formulae that are used here are:
According to the Bohr’s atomic model, the angular momentum of an electron present in a shell n is given by
$ L=\dfrac{nh}{2\pi } $
And the torque of an electron is given by
$ \begin{align}
& \tau =\dfrac{dL}{dt} \\
& \Rightarrow \tau =\dfrac{{{L}_{f}}-{{L}_{i}}}{{{t}_{f}}-{{t}_{i}}} \\
\end{align} $ .
Complete Step By Step Answer:
First of all, we must see that the time interval for the transition is already given i.e. $ 1.2\times {{10}^{-8}}s $ , now according to Bohr’s model of atom, the angular momentum for the various shells is given by
$ L=\dfrac{nh}{2\pi } $
For $ n=2 $ ,
$ {{L}_{2}}=\dfrac{2h}{2\pi }=\dfrac{h}{\pi } $
And for $ n=3 $
$ {{L}_{3}}=\dfrac{3h}{2\pi } $
Now as we know that the torque can be found from the angular momentum by the relation,
$ \tau =\dfrac{dL}{dt} $
i.e., for the case of transition of electron
$ \begin{align}
& \tau =\dfrac{{{L}_{2}}-{{L}_{3}}}{t} \\
& \Rightarrow \tau =\dfrac{\dfrac{3h}{2\pi }-\dfrac{2h}{2\pi }}{1.2\times {{10}^{-8}}} \\
& \Rightarrow \tau =\dfrac{\dfrac{h}{2\pi }}{1.2\times {{10}^{-8}}} \\
& \Rightarrow \tau =\dfrac{6.626\times {{10}^{-34}}}{2\times 3.14\times 1.2\times {{10}^{-8}}}=0.8792\times {{10}^{-26}} \\
& \Rightarrow \tau =8.79\times {{10}^{-27}} \\
\end{align} $
Hence, option $ \left( D \right)8.79\times {{10}^{-27}} $ is correct.
Note :
It is very important to see that the change in time $ dt $ given in the formula for the torque and angular momentum relation, is the time interval only, which is given in the question. The angular momentum of an electron in the atom depends directly on the shell in which the electron is present and the torque depends on the transition.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

What are the major means of transport Explain each class 12 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

Why should a magnesium ribbon be cleaned before burning class 12 chemistry CBSE

