Question

An electron in a hydrogen atom makes a transition from the first excited state to ground state. The equivalent current due to circulating electron:
A. Increases 2 times
B. Increases 4 times
C. Increases 8 times
D. Remains the same

Answer 1

Hint:The electric current is the rate of flow of charge through a cross-section per unit time. When the time taken to flow the same unit of charge increases then the magnitude of electric current decreases.

Formula used:
\[T = \left( {\dfrac{{4\varepsilon _0^4{h^3}}}{{m{e^4}}}} \right)\dfrac{{{n^3}}}{{{Z^2}}}\]
where T is the period of revolution of an electron in nth orbit of an atom of atomic number Z.
\[i = \dfrac{e}{T}\],
where I is the equivalent current due to the revolution of electrons in its orbit with period T.

Complete step by step solution:
When an electron is in orbit in any energy state of the atom, it crosses every point on the orbit after an interval of time called the period of revolution. So, the flow of electrons through a point per unit time will be equal to the magnitude of the equivalent current.

From the definition of electric current,
\[i = \dfrac{e}{T}\]
On putting the expression of the time period of revolution in nth state for electron, we get
\[i = \dfrac{e}{{\left( {\dfrac{{4\varepsilon _0^4{h^3}}}{{m{e^4}}}} \right)\left( {\dfrac{{{n^3}}}{{{Z^2}}}} \right)}} \\ \]
On simplifying the term and separating the constant and variable terms, we get
\[i = \left( {\dfrac{{m{e^5}{Z^2}}}{{4\varepsilon _0^4{h^3}}}} \right)\dfrac{1}{{{n^3}}} \\ \]
Here, Z is constant for a particular atom.

The electron is initially in its first excited state, \[{n_1} = 2\]. So, the equivalent current is,
\[{i_1} = \left( {\dfrac{{m{e^5}{Z^2}}}{{4\varepsilon _0^4{h^3}}}} \right)\dfrac{1}{{{2^3}}} \\ \]
\[\Rightarrow {i_1} = \left( {\dfrac{{m{e^5}{Z^2}}}{{4\varepsilon _0^4{h^3}}}} \right)\dfrac{1}{8} \\ \]
The electron is finally in ground state, \[{n_2} = 1\]. So, the equivalent current is,
\[{i_2} = \left( {\dfrac{{m{e^5}{Z^2}}}{{4\varepsilon _0^4{h^3}}}} \right)\dfrac{1}{{{1^3}}} \\ \]
\[\Rightarrow {i_2} = \left( {\dfrac{{m{e^5}{Z^2}}}{{4\varepsilon _0^4{h^3}}}} \right)\dfrac{1}{1}\]

Then the ratio of the current is,
\[\dfrac{{{i_1}}}{{{i_2}}} = \dfrac{{\left( {\dfrac{{m{e^5}{Z^2}}}{{4\varepsilon _0^4{h^3}}}} \right)\dfrac{1}{8}}}{{\left( {\dfrac{{m{e^5}{Z^2}}}{{4\varepsilon _0^4{h^3}}}} \right)\dfrac{1}{1}}} \\ \]
\[\Rightarrow \dfrac{{{i_1}}}{{{i_2}}} = \dfrac{1}{8} \\ \]
\[\therefore {i_2} = 8{i_1}\]
So, the final current is 8 times the initial current.

Therefore, the correct option is C.

Note: The orbit of the electron in an atom is assumed to be circular in nature. Then the electron is having uniform circular motion in its orbit and the angular momentum of the orbiting electron is always equal to the integral multiple of $\dfrac{h}{4\pi}$.