Question

The velocity of an electron in the nth orbit of a hydrogen atom bears the ratio 1:411 to velocity of light. The number of coloured lines formed when electron jumps from (n+3) state is:
(A) 4
(B) 3
(C) 5
(D) 6

Answer 1

Hint: Any arrangement of electrons that is higher in energy than the ground state: its energy is higher than the energy of the ground state. When an atom in an excited state undergoes a transition to the ground state in a process called decay, it loses energy by emitting a photon whose energy corresponds to the difference in energy between the two states.

Complete step by step solution: Velocity is defined as the rate of change of displacement with respect to time. It is classified as average velocity and instantaneous velocity.
The change of displacement in a certain amount of time is known as average velocity whereas the change of displacement at a particular instant of time is known as Instantaneous velocity.
Example is Bus moving with a velocity of 40m/s.
Relation between velocity and orbit is given by,
\[v=2165\times {{10}^{6}}\times \dfrac{z}{n}\]
Where n is the nth orbit.
\[\begin{align}
& \dfrac{{{v}_{n}}}{v}=\dfrac{1}{411} \\
& {{v}_{n}}=\dfrac{1}{411}\times 3\times {{10}^{8}}=7.299\times {{10}^{5}}m/s \\
\end{align}\]
\[\dfrac{7.299\times {{10}^{5}}}{2.165\times {{10}^{6}}}=\dfrac{1}{n}\]
Where Z=1 for hydrogen,
\[n=\dfrac{2.165\times {{10}^{6}}}{7.299\times {{10}^{5}}}=2.966=3\]
So hydrogen jumps from n=3 to n+3=6, total number of lines formed is given by,
\[\begin{align}
& lines=\dfrac{({{n}_{2}}-{{n}_{1}})({{n}_{2}}-{{n}_{1}}+1)}{2}=\dfrac{(6-3)(6-3+1)}{2} \\
& lines=\dfrac{3\times 4}{2}=6 \\
\end{align}\]

So, the answer is option (D).

Note: The energy levels of a hydrogen atom had to be quantized; in other words, only states that had certain values of energy were possible, or allowed. If a hydrogen atom could have any value of energy, then a continuous spectrum would have been observed, similar to blackbody radiation.