Question

An atom typically exists in an excited state for about $\Delta t={10}^{-8}$ s. Estimate the uncertainity in frequency $\Delta \nu$ in the frequency of emitted photons when an atom makes a transition from an excited state with the simultaneous emission of a photon with an average frequency of ${\nu }_{av}$ = $7.1\times {10}^{14}$ Hz. Calculate the ratio of uncertainity in frequency to the average frequency.

• A. $1.12\times {10}^{-10}$
• B. $2.11\times {10}^{-10}$
• C. $2.11\times {10}^{-8}$
• D. $1.12\times {10}^{-8}$

Answer 1

The correct option is A $1.12\times {10}^{-10}$

We know,
From Heisenberg uncertainty principle,
$\Delta E.\left(\Delta t\right)\ge \frac{h}{4\pi }$
Again, from Plack's quantum theory, we can say $E=h\Delta \nu$
Combining the relations, we get, $h(\Delta .\nu )\Delta t\ge \frac{h}{4\pi }$
Or $\left(\Delta \nu \right)\ge \frac{h}{4\pi .\Delta t.h}$
$\Rightarrow \left(\Delta \nu \right)$ = $\frac{1}{4\pi \times {10}^{-8}}\approx 8\times {10}^6$ Hz
So, $\frac{\Delta \nu }{{\nu }_{av}}$ = $\frac{8\times {10}^6}{7.1\times {10}^{14}}$ = $1.12\times {10}^{-8}$