Question

The energy required to remove an electron from $ n = 2 $ state in hydrogen will be
A) $ 13.6eV $
B) $ 3.4eV $
C) $ 27.2eV $
D) $ 6.8eV $

Answer 1

Hint: Hydrogen is a chemical element with atomic number $ 1 $ . The energy in $ {n^{th}} $ level can be calculated from the energy in ground state and atomic number. The energy in the $ {2^{nd}} $ energy level can be calculated from the below formula, as the given energy level is $ n = 2 $ .
 $ {E_n} = \dfrac{{{E_0} \times {Z^2}}}{{{n^2}}} $
 $ {E_n} $ is energy of $ {n^{th}} $ level
 $ {E_0} $ is energy in ground state
Z is atomic number
n is the number of energy levels.

Complete answer:
Given chemical element is hydrogen. Hydrogen is an element with atomic number $ 1 $ . It has only one electron in its shell or orbital.
The energy of hydrogen in its ground state is $ {E_0} = - 13.6eV $
The atomic number of hydrogen is $ Z = 1 $
We have to calculate the energy required to remove an electron from the $ n = 2 $ state of hydrogen.
Here, $ n = 2 $ as the electron is removed from the second energy level.
By substituting the above values in the formula, we will get
 $ {E_2} = \dfrac{{ - 13.6 \times {1^2}}}{{{2^2}}} $
By solving the above calculation, the result will be
 $ {E_2} = - 3.4eV $
The energy was expressed in units of electron volts. It can also be expressed in joules by multiplying with the $ 1.6 \times {10^{ - 19}} $ .
The given options are in the units of electron volts.
Option B is the correct one.

Note:
The lowest energy can always be called the ground state and the higher levels can be called the excited state. Bohr introduced the concept of energy levels and successful in finding out the energies of hydrogen atom and similar ions like helium ion $ \left( {H{e^ + }} \right) $ and lithium ion $ \left( {L{i^{2 + }}} \right) $ .