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Question: In a hydrogen atom, an electron makes transition from \( n=3 \) to \( n=1 \) state in time interval ...

In a hydrogen atom, an electron makes transition from n=3n=3 to n=1n=1 state in time interval of 1.2×10−8s1.2\times {{10}^{-8}}s , calculate average torque (Nm)\left( Nm \right) acting on the electron during this transition.
(A)1.055×10−26 (B)4.40×10−27 (C)1.7×10−26 (D)8.79×10−27 \begin{aligned} & \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{aligned}

Explanation

Solution

Hint : In order to solve this question, we are going to first calculate the angular momentum of the electron in both the shells 22 and 33 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=nh2πL=\dfrac{nh}{2\pi }
And the torque of an electron is given by
τ=dLdt ⇒τ=Lf−Litf−ti \begin{aligned} & \tau =\dfrac{dL}{dt} \\\ & \Rightarrow \tau =\dfrac{{{L}_{f}}-{{L}_{i}}}{{{t}_{f}}-{{t}_{i}}} \\\ \end{aligned} .

Complete Step By Step Answer:
First of all, we must see that the time interval for the transition is already given i.e. 1.2×10−8s1.2\times {{10}^{-8}}s , now according to Bohr’s model of atom, the angular momentum for the various shells is given by
L=nh2πL=\dfrac{nh}{2\pi }
For n=2n=2 ,
L2=2h2π=hπ{{L}_{2}}=\dfrac{2h}{2\pi }=\dfrac{h}{\pi }
And for n=3n=3
L3=3h2π{{L}_{3}}=\dfrac{3h}{2\pi }
Now as we know that the torque can be found from the angular momentum by the relation,
τ=dLdt\tau =\dfrac{dL}{dt}
i.e., for the case of transition of electron
τ=L2−L3t ⇒τ=3h2π−2h2π1.2×10−8 ⇒τ=h2π1.2×10−8 ⇒τ=6.626×10−342×3.14×1.2×10−8=0.8792×10−26 ⇒τ=8.79×10−27 \begin{aligned} & \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{aligned}
Hence, option (D)8.79×10−27\left( D \right)8.79\times {{10}^{-27}} is correct.

Note :
It is very important to see that the change in time dtdt 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.