(a) State Bohr’s postulate to define stable orbits in hydrogen atom. How does de Broglie’s hypothesis explain the stability of these orbits?
(b) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon.

 

 

Answers (1)
S safeer

Bohr's second postulate defines stable orbits. The postulate states that "The electron revolves around the nucleus only in those orbits for which its  angular is an integral multiple of \frac{h}{2\pi }"
That is L = angular momentum
           h = plank's constant
According to de Broglie's hypothesis, material particles such that electron also have wave nature.
For an electron moving in the nth circular orbit of radius rn, the total distance is the circumference of the orbit
Thus

n\lambda = 2\pi r_{n}\: \: ; n= 1,2,3\cdots
\lambda = de Broglie wavelength of an electron moving in nth orbit
\lambda = \frac{h}{p}
if the speed of the electron is much less than the speed of light then,
p= mv_{n}
\lambda = \frac{h}{mv_{n}}, \frac{nh}{mv_{n}}= 2\pi r_{n}
mv_{n}r_{n}= \frac{nh}{2\pi }= L
This is the condition proposed by Bohr for the angular momentum of electron.
b) The energy of nth level of the Hydrogen atom is
E_{n}= \frac{-13\cdot 6}{n^{2}}eV
here hydrogen atom is excited from n = 1 to n = 4 so the energy of the photon will be
E_{4}-E_{1}= h\vartheta
where \vartheta is the frequency of the absorbed photon
Therefore, we can write
\frac{-13\cdot 6}{4^{2}}-\left ( -13\cdot 6 \right )= \frac{6\cdot 63\times 10^{34}\times \vartheta }{1\cdot 6\times 10^{-19}}
\vartheta = \frac{13\cdot 6\times 16\times 10^{-19}\times15}{16\times 6\cdot 63\times 10^{-34}}= 3\cdot 1\times 10^{15}\, H{z}
Therefore the frequency of the absorbed photon is 3\cdot 1\times 10^{-15}\, H{z}
 

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