Question

Let \(T_1\) and \(T_2\) be the energy of an electron in the first and second excited states of the hydrogen atom, respectively. According to Bohr’s model of an atom; the ratio T1:T2 is:

• A. \(1:4\)
• B. \(4:1\)
• C. \(4:9\)
• D. \(9:4\)

Answer 1

The Correct Option is D

To determine the ratio of the energy levels of an electron in different states of a hydrogen atom according to Bohr’s model, we start by understanding how the energy of an electron in a given state is defined. According to Bohr's model, the energy of an electron in the \(n^{th}\) orbit is given by the formula:

\(E_n = -\dfrac{13.6}{n^2}\, \text{eV}\)
where \(E_n\) is the energy in electron volts (eV) and \(n\) is the principal quantum number corresponding to the orbit.

We need to find the energies \(T_1\) for the first excited state and \(T_2\) for the second excited state of the hydrogen atom.
For the first excited state, \(n = 2\):

\(T_1 = -\dfrac{13.6}{2^2} = -\dfrac{13.6}{4}\, \text{eV}\)
For the second excited state, \(n = 3\):

\(T_2 = -\dfrac{13.6}{3^2} = -\dfrac{13.6}{9}\, \text{eV}\)

The absolute value of these energies is used to find the ratio since energy values are negative due to the nature of bound states.

The ratio \(\dfrac{T_1}{T_2}\) is:
 

\(\dfrac{\left|\dfrac{13.6}{4}\right|}{\left|\dfrac{13.6}{9}\right|} =  \dfrac{9}{4}\)

Thus, the ratio of the energies \(T_1:T_2\) is: \(9:4\)