atoms Question 23
Question: Q. 1. (i) State Bohr’s postulate to define stable orbits in hydrogen atom. How does de-Broglie’s hypothesis explain stability of these orbits ?
(ii) 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.
R [CBSE 2017, 2018]
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Solution:
Ans. (i) Statement of Bohr’s postulate Explanation in terms of de-Broglie hypothesis $1 / 2$
(ii) Finding the energy in the $n=4$ level
Estimating the frequency of the photon
(i) Bohr’s postulate, for stable orbits, states “The electron, in an atom, revolves around the nucleus only in those orbits for which its angular momentum is an integral multiple of $\frac{h}{2 \pi}(h=$ Planck’s constant),” $1 / 2$ [Also accept $m v r=n \cdot \frac{h}{2 \pi} \quad(n=1,2,3, \ldots \ldots)$
As per de-Broglie’s hypothesis, $\lambda=\frac{h}{p}=\frac{h}{m v}$
For a stable orbit, we must have circumference of the orbit $=n \lambda(n=1,2,3, \ldots \ldots$.
$$ \begin{align*} \therefore & 2 \pi r & =n \cdot m v \ \text { or } & m v r & =\frac{n h}{2 \pi} \end{align*} $$
Thus de-Broglie showed that formation of stationary pattern for integral ’ $n$ ’ gives rise to stability of the atom.
This is nothing but the Bohr’s postulate.
$1 / 2$ (ii) Energy in the $n=4$ level $=\frac{-E_{0}}{4^{2}}=-\frac{E_{0}}{16}$
$\therefore$ Energy required to take the electron from the ground state, to the $n=4$ level $=\left(-\frac{E_{0}}{16}\right)-\left(-E_{0}\right)$
$=\frac{-1+16}{16} E_{0}=\frac{15}{16} E_{0}=\frac{15}{16} \times 13.6 \times 1.6 \times 10^{-19} \mathrm{~J} \quad 1 / 2$
Let the frequency of the photon be $v$, we have
$$ \begin{array}{rlrl} \quad h v & =\frac{15}{16} \times 13.6 \times 1.6 \times 10^{-19} \ \therefore \quad & v & =\frac{15 \times 13.6 \times 1.6 \times 10^{-19}}{16 \times 6.63 \times 10^{-34}} \mathrm{~Hz} \ & \left.\simeq 3.1 \times 10^{15} \mathrm{~Hz} \text { (Also accept } 3 \times 10^{15} \mathrm{~Hz}\right) \end{array} $$
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