Q. 9

# State Bohr’s quantization condition of angular momentum. Calculate the shortest wavelength of the Brackett series and state to which part of the electromagnetic spectrum does it belong.

OR

Calculate the orbital period of the electron in the first excited state of hydrogen atom.

Answer :

__Bohr’s quantization condition:__ The angular momentum of an electron in an orbit around the hydrogen atom has to be an integral multiple of Planck’s constant divided by twice π. Mathematically,

The wavelength can be calculated in case of Brackett series by the formula:

For the wavelength to be shortest, n has to be maximum i.e. ∞ .

Hence,

It is between 700 nm and 1mm which is the range of infrared. Hence, it belongs to the infrared part of the spectrum.

OR

Let the radius of the shell of the first exited state be r and the speed of electron in the second exited stated be v.

Note that the ground state is and the first exited state is .

The angular momentum of the electron in the first exited state (n=2) is

Now, the force on the electron is given by coulomb’s law:

where is the magnitude of charge on electron.

The centripetal force on electron is:

The coulomb force acts as centripetal force. Hence,

Substituting the value of v from (1) in (2).

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Suppose in an imaginary world the angular momentum is quantized to be even integral multiples of h/2π. What is the longest possible wavelength emitted by hydrogen atoms in visible range in such a world according to Bohr’s model?

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