Question

A photon of wavelength 300nm interacts with a stationary hydrogen atom in ground state. During the interaction, the whole energy of the photon is transferred to the electron of the atom. State which possibility is correct.
(Consider, ${{E}_{0}}=\dfrac{m{{k}^{2}}e{}^{4}}{2{{h}^{2}}{{n}^{2}}}=\dfrac{k{{e}^{2}}}{2{{a}_{o}}}=13.6eVs$ velocity of light = \[3\times {{10}^{8}}m/s\], ionization energy of hydrogen = 13.6eVs)
A. Electron will be knocked out of atom
B. Electron will go to any excited state of the atom
C. Electron will go only to first excited state of an atom
D. Electron will keep orbiting on the ground state of the atom

Answer 1

Hint: You can pick keywords from which you can proceed to solve a given question using given data. Like photo energy, electron energy, ground state of hydrogen, hydrogen is steady etc. use formulas of above terms to get the desired result.

Complete Step-by-Step solution:
In the question, we have a wavelength of photons 300 nm. A photon interacts with hydrogen. This hydrogen is a steady one in ground level.
$\lambda =330nm$
During this interaction the whole energy of the photon is transferred to the electron of an atom. Then we have to calculate what will happen to the electron and what will be its behavior.
Aim to calculate: what will happen to the electron and what will be its behavior?
To solve this, first calculate the energy of the photon. As in the question they mention word photon energy,
We know that,
$E{}_{photon}=hv$
And $\begin{align}
  & v=\dfrac{c}{\lambda } \\
 & therefore, \\
 & {{E}_{photon}}=\dfrac{4\times {{10}^{-15}}\times 3\times {{10}^{8}}}{300\times {{10}^{-9}}} \\
\end{align}$
By solving,
$E{}_{photon}=4eV$ ---------(1)
Hydrogen is at ground level. We know that the value of energy of hydrogen is 13.6 $eV$.
${{E}_{hydrogen ground state}}=13.6eV$ ---------(2)
As you can see in the equation one and two, the energy of a photon is less than the energy of an electron. So the minimum energy required to excite in the next state of hydrogen atom i.e at n=2 must be greater than 4$eV$. Hence no excitation takes place and electrons will remain in orbit on the ground state of the atom.
Correct Option is (D)

Note: Focus on energy of hydrogen. Formula to calculate energy of ground level of hydrogen is given below,
${{E}_{0}}=\dfrac{m{{k}^{2}}e{}^{4}}{2{{h}^{2}}{{n}^{2}}}=\dfrac{k{{e}^{2}}}{2{{a}_{o}}}=13.6eVs$
For the ground state of a hydrogen atom, the value of the quantum number (n) must be 1. Do not put a value of n=0 for ground level.