Question

The ratio of $ \left( {{{\text{E}}_{\text{2}}}{\text{ - }}{{\text{E}}_{\text{1}}}} \right) $ to $ \left( {{{\text{E}}_{\text{4}}}{\text{ - }}{{\text{E}}_{\text{3}}}} \right) $ for H-atom is approximately:
(A) 10.2
(B) 15.4
(C) 5.6
(D) 12.4

Answer 1

Hint: Bohr’s model of the atom says that the energy corresponding to a certain energy level of an atom is inversely proportional to the square of the number of that energy level. The energy associated with the first four orbits of the hydrogen atom can be evaluated from this theory and the required ratio can be calculated.

Complete step by step solution:
According to the Bohr’s model of the atom, an atom is made up of a large positively charged nucleus with the electrons revolving around it in fixed circular orbits called energy levels of the atom. Each of these energy levels or orbits is associated with a definite amount of energy.
With the increase in the distance from the nucleus, the energy associated with different levels of energy also increases. These energy levels are represented by the letters K, L, M, N etc. or by the numbers 1, 2, 3, 4 etc.
According to Bohr’s theory, the energy associated with an energy level of an atom is given by the expression:
$E_{n}= -\dfrac{2\Pi ^{2}me^{4}Z^{2}}{n^{2}h^{2}}$
where $ {{\text{E}}_{\text{n}}} $ represents the energy associated with the nth energy level, Z is the atomic number of the atom, m denotes the mass of the electron, e denotes the charge of the electron and h is the Planck’s constant which is equal to $ 6.626\times 10^{-34}Js $ .
Now, we have mass of electron, $ m= 9.1\times 10^{-31}kg $ and charge of electron, $ e= 1.602\times 10^{-19}C $ .
The expression can be rewritten by substituting these values:
 $ {{\text{E}}_{\text{n}}}{\text{ = - }}\dfrac{{{\text{13}}{\text{.6}}{{\text{Z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}} $
For H- atoms, the atomic number is 1. So $ {\text{Z = 1}} $ . Therefore we will now have:
 $ {{\text{E}}_{\text{n}}}{\text{ = - }}\dfrac{{{\text{13}}{\text{.6}}}}{{{{\text{n}}^{\text{2}}}}} $
This is the relation for the energy associated with the nth orbit of the H- atom.
For the first orbit, n is equal to 1 and we have $ {{\text{E}}_{\text{1}}}{\text{ = - }}\dfrac{{{\text{13}}{\text{.6}}}}{{{{\text{1}}^{\text{2}}}}}{\text{ = - 13}}{\text{.6}} $
For the second orbit, n is equal to 2 and we have $ {{\text{E}}_{\text{2}}}{\text{ = - }}\dfrac{{{\text{13}}{\text{.6}}}}{{{{\text{2}}^{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{1}}}}}{{\text{4}}} $
For the third orbit, n is equal to 3 and we have $ {{\text{E}}_{\text{3}}}{\text{ = - }}\dfrac{{{\text{13}}{\text{.6}}}}{{{{\text{3}}^{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{1}}}}}{{\text{9}}} $
For the fourth orbit, n is equal to 4 and we have $ {{\text{E}}_{\text{4}}}{\text{ = - }}\dfrac{{{\text{13}}{\text{.6}}}}{{{{\text{4}}^{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{E}}_{\text{1}}}}}{{{\text{16}}}} $
Therefore, $ $
  $
E_{2}-E_{1}= \dfrac{E_{1}}{4}-E_{1}
\Rightarrow E_{2}-E_{1}= -\dfrac{3E_{1}}{4}
 $
and
 $
E_{4}-E_{3}= \dfrac{E_{1}}{16}-\dfrac{E_{1}}{9}
\Rightarrow E_{4}-E_{3}= -\dfrac{7E_{1}}{16\times 9}
 $
Therefore, the required ratio $ \left( {{{\text{E}}_{\text{2}}}{\text{ - }}{{\text{E}}_{\text{1}}}} \right) $ to $ \left( {{{\text{E}}_{\text{4}}}{\text{ - }}{{\text{E}}_{\text{3}}}} \right) $ for H-atom is approximately:
  $
  \dfrac{E_{2}-E_{1}}{E_{4}-E_{3}}= \dfrac{\dfrac{-3E_{1}}{4}}{\dfrac{-7E_{1}}{16\times 9}}
\Rightarrow \dfrac{E_{2}-E_{1}}{E_{4}-E_{3}}= \dfrac{3\times 16\times 9}{7\times 4}
\Rightarrow \dfrac{E_{2}-E_{1}}{E_{4}-E_{3}}= 15.4
  $
So the correct option is B.

Note:
The energy difference between any two successive energy levels of an atom is not the same. With increase in the value of n, it goes on decreasing. The energy is emitted or absorbed discontinuously in the form of small packets called quantas only during electron transitions between two stationary states. The energy difference between these two states gives the frequency of the light.