Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Calculate‌ ‌the‌ ‌radius‌ ‌of‌ ‌the‌ ‌Bohr’s‌ ‌first‌ ‌orbit‌ ‌for‌ ‌hydrogen‌ ‌atom‌ ‌and‌ ‌the‌ ‌energy‌ ‌of‌ ‌electron‌ ‌in‌ ‌this‌ ‌orbit:‌ ‌A)$‌ ‌-‌ ‌2.178‌ ‌\times‌ ‌{10^{‌ ‌-‌ ‌18}}J$‌ ‌B)$21.78‌ ‌\times‌ ‌{10^{‌ ‌-‌ ‌18}}J$‌ ‌C)$2.170‌ ‌\times‌ ‌{10^{‌ ‌-‌ ‌18}}J$‌ ‌D)$21.88‌ ‌\times‌ ‌{10^{‌ ‌-‌ ‌18}}J$‌

Last updated date: 04th Aug 2024
Total views: 355.2k
Views today: 3.55k
Verified
355.2k+ views
Hint: According to Bohr's theory the angular momentum is an integral of Planck’s constant divided by twice the value of pi. Bohr’s theory was only valid for one electron system and then the electron revolved around the nucleus using the centripetal force.

Bohr’s theory was based on the application of Planck’s quantum theory of the atomic spectra of hydrogen atoms.
The postulates of the theory are: -
The electron in an atom has only certain definite stationary states which are called as energy levels.in each of the energy levels the electron revolves in a circular motion, due to the centripetal force acting on the electron from the centre.
Only those states are allowed electronic motion which have the angular momentum of an electron as an integral multiple of $\dfrac{h}{{2\pi }}$.
Now as we know that the force on the electron is centripetal, therefore the electrostatic force between the electron and nucleus is
Let the charge on the electron be ‘e’ and that will be the same charge on the nucleus due to just one proton. Also, the mass of electron is defined as ${m_e}$
$\dfrac{{KZ{e^2}}}{{{r^2}}} = \dfrac{{{m_e}{v^2}}}{r}$ …. Eq (1)
Also, we know that only those states are allowed that have the angular momentum as an integral multiple of $\dfrac{h}{{2\pi }}$
Therefore
${m_e}vr = n\dfrac{h}{{2\pi }}$ …… Eq (2)
Solving equation one and two, we get
$r = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}m{e^2}KZ}} \\ \\$
Putting the values of the above given quantities, we get
$n = 1 \\ h = 6.626 \times {10^{ - 35}}Js \\ K = 9 \times {10^9}N{m^2}{C^{ - 2}} \\ e = 1.6 \times {10^{ - 19}}C \\$
We get
$r = 0.529 \times {10^{ - 10}}\dfrac{{{n^2}}}{Z}$
The hydrogen atom is hydrogen therefore the atomic number is one and as we need to find out the first energy state therefore ‘n’ is also one.
$r = 0.529 \times {10^{ - 10}}m$
Now the energy can be defined as the energy needed to take out one electron completely out of protons reach, now for that energy can be written as
$E = - \dfrac{{KZ{e^2}}}{{2r}}$
Putting the values, we get
$E = - 2.178 \times {10^{ - 18}}J \times \dfrac{{{Z^2}}}{{{n^2}}}$
Here the atomic number is one and so is the energy state, therefore the energy is equal to $E = - 2.178 \times {10^{ - 18}}J$
For the first energy state of a boar hydrogen atom, the radius is $r = 0.529 \times {10^{ - 10}}\dfrac{{{n^2}}}{Z}$and the energy $E = - 2.178 \times {10^{ - 18}}J$.

Note:
For a Bohr atom the radius is directly proportional to the energy state and inversely proportional to the atomic number, while energy is directly proportional to the atomic number and inversely proportional to the energy state.